Mathematics-22.Unit-18-Ellipse and Hyperbola
ELLIPSE
SYLLABUS
Equations of ellipse and hyperbola in standard form, their foci, directrices and eccentricity, parametric equations, equations of tangent and normal.
ELLIPSE
DEFINITION
An ellipse is the locus of a point which moves in a plane such that the ratio of its distances from a fixed point (called focus) and from a fixed straight line (called directrix) is always constant and less than 1. And this constant ratio is called the eccentricity of the ellipse.
1.1 STANDARD EQUATION OF THE ELLIPSE
+ = 1 or ,
where b2 = a2(1 - e2).
The eccentricity of the ellipse is given by the relation b2 = a2(1 - e2), i.e., e2 = 1 - b2/a2
An ellipse has two foci and two directrices.
Latus Rectum: Latus rectum is the line which passes through the focus of the ellipse and perpendicular to the major axis. End points of the latus rectum are given by(ae, b2/a) and (ae,-b2/a).Also the length of semi latus rectum is given by b2 / a.
Focal Distance of a Point: To find the distance of any point on the ellipse from the focus, we use the definition of the ellipse.
Let P(x, y) be a point on the ellipse. Then
SP = ePN = e(a/e –x) = a - ex
SP = ePN = e ( a/e + x) = a + ex
SP + SP = 2a
the sum of the focal distances of any point on the ellipse is equal to its major axis. Also SS’ b
Focus (X, Y) is (ae, 0) and (–ae, 0)
Þ (x, y) for focus Þ (1 + ae, 2) and (1 – ae, 2)
Þ foci are (1 + , 2) and (1 – , 2)
Similarly equation of directrices are X = and X = –
Þ x = 1 + and x = 1 – are the directrices.
AUXILIARY CIRCLE
The circle described on the major axis of an ellipse as diameter is called the Auxiliary Circle of the ellipse. Equation of the auxiliary circle is x2+y2=a2.
PARAMETRIC EQUATION OF THE ELLIPSE
Thus for any point P(x, y) on the ellipse we have x = a cos, y = b sin are called parametric equation of the ellipse.
Position of a Point Relative to an Ellipse:
The point P(x1, y1) is outside or inside or on the ellipse according as the quantity
S1 º is positive or negative or zero.
Illustration 5: Consider the ellipse x2 + 3y2 = 6 and a point P on it in the first quadrant at a distance of 2 units from the centre. Find the eccentric angle of P.
Solution: Equation of ellipse is x2 + 3y2 = 6
Equation of auxiliary circle is x2 + y2 = 6
Since P (x1, y1) & Q (x1, y2) lie on the ellipse and the circle respectively we have,
x12 + 3y12 = 6 …..(1)
x12 + y22 = 6 …..(2)
3y12 –y22 = 0
Again OP = 2 x12 + y12 = 4 …..(3)
By (1) –(3), we get,
2y12 = 2 y12 = 1 y1 = 1 [ P is in the first quadrant]
y2 =
Putting y1 in (1), we get x12 = 3 x1 =
Eccentric angle of P = =
EQUATION OF THE TANGENT AT A POINT OF AN ELLIPSE
(i) Let the equation of ellipse be .
Slope of tangent to the ellipse at a point (x1, y1) =
Hence the equation of the tangent at (x1, y1) is y -y1 =
i.e. T= 0
(ii) Equation of tangent at the point i.e. (a cos, b sin) is obtained by putting
x1 = a cos, y1 = b sin;
EQUATION OF THE TANGENT IN TERMS OF ITS SLOPE; USING THE CONCEPT OF COMPARISON:
The equations of the tangent to the ellipse =1 with slope m are
y = mx for all finite values of m.
Moreover the line touches the ellipse .
Illustration 6: From a point P two tangents are drawn one each to the ellipse , If tangents are perpendicular to each other, then find locus of P.
Solution: The tangent at (a cosq, b sinq) on is
….(i)
The tangent at (a cosf, b sinf) on is
….(ii)
(i) and (ii) are perpendicular cos (q - f) = 0 or q = f +
….(iii)
Eliminating f from (ii) and (iii)
Locus is (x2 + y2)2 = b2 (x + y)2 + a2(x - y)2
EQUATION OF THE PAIR OF TANGENTS DRAWN FROM THE POINT (X1, Y1)
The following equation will be the equation of the pair of tangents from (x1, y1).
If we assume that T = – 1, S = , and
S1 = , then equation becomes
(T – S1)2 = S1(S – 2T + S1)
i.e. T2 = SS1.
Illustration 7: Find the angle between the pair of tangents drawn to the ellipse 3x2 + 2y2 = 5 from the point (1, 2).
Solution: Let the equation of the line passing through (1, 2) be
y – 2 = m (x – 1) or y = mx – m + 2 … (1)
Line (1) touches 3x2 + 2y2 = 5 if
3x2 + 2(mx – m + 2)2 = 5
or (3 + 2m2)x2 – 4m(m–2)x + 2m2 + 8 – 8m – 5 = 0
or (3 + 2m2)x2 – 4m(m–2)x + 2m2 – 8m +3 = 0
For equal roots. D = 0
[4m(m–2)]2 – 4(3 + 2m2)(2m2 – 8m + 3) = 0
or 4m2 (m2 – 4m + 4) – (6m2 – 24m + 9 + 4m4 – 16m3 + 6m2) = 0
or 4m4 – 16m3 + 16m2 –12m2 – 4m4 + 16m3 + 24m – 9 = 0
or 4m2 + 24m – 9 = 0
m1 + m2 = –6 and m1m2 = –9/4
tan =
= tan–1
Alternative Method 1:
Hint: Find the equation of pair of tangent i.e. T2 = SS1,
Find the angle between above two tangents i.e. tan =
Alternative Method 2:
Equation to the ellipse is
Let the equation to tangent be y = mx +
Which is passing through (1, 2) 2 = m +
(2 – m)2 = 4m2 + 24m – 9 = 0 has two roots m1 and m2 which are the slopes of the tangents from the point (1, 2)
m1 + m2 = –6 ; m1m2 = –9/4
Angle between the tangents is tan–1
Director Circle
The locus of the point of intersection of a pair of perpendicular tangents to an ellipse is called Director Circle.
EQUATION OF DIRECTOR CIRCLE
x2 + y2 = a2 + b2
EQUATION OF THE NORMAL AT A POINT OF AN ELLIPSE
Equation of the normal is y-y1 =
• Equation of normal at (a cos, b sin) is = a2 – b2
ax sec - by cosec = a2 - b2
Example -8: If the normals to the ellipse at the points (x1, y1), (x2, y2) and (x3, y3) are concurrent, prove that = 0.
Solution: The equation of the normal to the given ellipse at (x1, y1) is
a2xy1 – b2yx1 – (a2 – b2)x1y1 = 0 . . .(1)
Similarly the normals at (x2, y2) and (x3, y3) are
a2xy2 – b2yx2 – (a2 – b2)x2y2 = 0 …(2)
a2xy3 – b2yx3 – (a2 – b2)x3y3 = 0 …(3)
Eliminating a2x, b2y and (a2 - b2) from (1), (2) and (3), we find that the three lines are concurrent if
Þ = 0 Þ = 0
Illustration9: Prove that the straight line lx + my = n is a normal to the ellipse, = 1 if .
Solution: lx + my = n must be of the form
ax sec by cosec = a2 b2
so, comparing we get,
so, cos = , sin = , squaring and adding
we get,
EQUATION OF MID POINT CHORD
Equation of mid chord whose mid point is (x1,y1) is given by
T=S1 where T= and S1= .
Illustration 10: Find the locus of the midpoint of chords of the ellipse , that are parallel to the line y = 2x + c.
Solution: Let the mid – point of the chord be (h, k). Then equation of this chord will be
T = S1
Slope of line
Required locus is
Illustration 11: Find the locus of the middle point of chord of the ellipse = 1, which are drawn through the positive end of the minor axis.
Solution: Let the middle point be (h, k), so equation is T = S1
. As it passes through (0, b)
so, , so locus is
CHORD OF CONTACT
Equation of chord of contact is given by T = 0, where T = .
Illustration 12: From a point O on the circle x2 + y2 = 25, tangents OP and OQ are drawn to the ellipse Show that the locus of the mid point of the chord PQ describes the curve x2 + y2 = 25 .
Solution: O (5 cos, 5 sin)
Let R be middle point of PQ. Let R (h, k) .
Equation of the chord of contact PQ is T = 0
i.e. . . . . . (1)
Equation of the chord PQ with middle point R(h, k) is
T = S1 i.e.
or . . . .(2)
(1) and (2) are the same equations.
= 1
Hence locus of (h, k) is (x2 +y2) = 25 .
Hyperbola
DEFINITION
A hyperbola is the locus of a point which moves such that, ratio of its distance from a fixed point (focus) and its distance from a fixed straight line (directrix), is a constant (eccentricity). This constant (eccentricity) is greater than unity.
1.1 STANDARD EQUATION AND BASIC DEFINITIONS
.
(i) The eccentricity e is given by the relation
(ii) Since the curve is symmetrical about the y - axis, it is clear that there exists another focus S at (-ae, 0) and a corresponding directirx ZM with the equation x= -a/e
(iii) The points A and A are called the vertices of the hyperbola.
(iv) The straight line joining the vertices is called the transverse axis of the hyperbola, its length AA is 2a.
(v) The straight line BB is called the conjugate axis.
Illustration 13: Find the eccentricity of the hyperbola which passes through (3, 0) and (3 , 2).
Solution: Let the hyperbola be
It passes through (3, 0) and (3 , 2)
and = 1
Which give a2 = 9 and b2 = 4
from b2 = a2(e2 –1), we get 4 = 9(e2 –1)
or e2 = or e = .
Illustration 14: Find the equation of the hyperbola whose directrix is 2x + y =1, focus (1, 2) and eccentricity .
Solution: Let S (1, 2) be the focus and P (x, y) be a point on the hyperbola. Draw PM perpendicular form P on the directrix. Then by definition
SP = ePM
(x –1)2 + (y –2)2 = 3
5{(x –1)2 + (y –2)2} = 3{2x + y –1}2
5x2 + 5y2 –10x –20y + 25 = 3(4x2 + y2 + 1 + 4xy –4x –2y)
7x2 –2y2 + 12xy –2x + 14y –22 = 0
This is the required equation of the hyperbola.
1.2 RELATION BETWEEN FOCAL DISTANCES
The difference of the focal distances of a point on the hyperbola is constant. PM and PM are perpendiculars to the directrices MZ and MZ
PS - PS = e(PM - PM)
= eMM = e(2a/e) = 2a = constant.
1.3 RELATIVE POSITION OF A POINT WITH RESPECT TO THE HYPERBOLA
The quantity is positive, zero or negative, according as the point (x1, y1) lies within, upon or without the curve.
1.4 PARAMETRIC COORDINATES
Thus Any point on the curve, in parametric form is
X = a sec, y = b tan.
In other words, (a sec, b tan ) is a point on the hyperbola for all values of . The point (a sec, b tan) is briefly called the point ''.
1.5 IMPORTANT PROPERTIES OF HYPERBOLA
Since the fundamental equation of the hyperbola only differs from that of the ellipse in having -b2 instead of b2, it will be found that many propositions for the hyperbola are derived from those for the ellipse by changing the sign of b2. Some results for the hyperbola are
(i) The tangent at any point (x1, y1) on the curve is -
(ii) The tangent at the point ‘’ is
(iii) The straight line y = mx + c is a tangent to the curve, if c2 = a2 m2 – b2. In other words,
y = mx touches the curve, for all those values of m when m > b/a or m< –b/a.
(iv) The straight line lx + my = n is a tangent to the hyperbola = 1 if n2 = a2l2 – b2m2.
(v) Equation of the normal at any point (x1, y1) to the curve is =
(vi) The equation of the chord through the points 1 and 2 is
(vii)
(viii) The equation of the normal at is ax cos + by cot = a2 + b2
(ix) Through a given point, four normals can be drawn to a hyperbola (real or imaginary).
(x) Tangent drawn at any point bisects the angle between the lines, joining the point to the foci , whereas normal bisects the supplementary angle between the lines.
(xi) Equation of director circle is x2 + y2 = a2– b2. That means if a2 > b2, there would exist several points such that tangents drawn from them would be mutually perpendicular. If
a2 < b2, no such point exist. For a2 = b2, centre is the only point from which two perpendicular tangents (asymptotes) to the hyperbola can be drawn.
(xii) A straight line x cos + y sin = p is tangent if p2 = a2 cos2 – b2 sin2 .
(xiii) Equation of straight line passing through the point (a sec, b tan) and (a sec , b tan ) is .
Illustration15: Tangents are drawn from point P on the curve x2 – 4y2 = 4 to the curve x2 + 4y2 = 4 touching it in the points Q and R. Prove that the mid – point of QR lies on
Solution: Any point on the curve x2 – 4y2 = 4 can be expressed as ( 2 secq, tanq)
Let (h, k) be the mid – point of QR which is a chord to the curve x2 + 4y2 = 4
Equation of QR is
T= S1
QR is also the chord of contact to the curve x2 + 4y2 = 4 w.r.t the point ( 2 sec q, tanq)
\ Equation of chord of contact of (2 sec q, tan q) w.r.t x2 + 4y2 = 4 is
secq . +tanq y =1
Comparing the two equation
secq =
Eliminating q sec2 q – tan2 q =1
\ Required locus of (h, k) is
CONJUGATE HYPERBOLA
The hyperbola, whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola, is called the conjugate hyperbola of the given hyperbola, and the two hyperbolas are conjugate to one another. Thus, the hyperbolas and are conjugate hyperbolas.
Illustration 16: Tangents are drawn to a hyperbola from any point on one of the branches of the conjugate hyperbola. Show that their chord of contact will touch the other branch of the conjugate hyperbola.
Solution: Let the hyperbola be = 1. So its conjugate hyperbola is = 1. Let any point on it be (a tan , b sec ). Now equation of chord of contact will be
tan – sec = 1 (–tan ) – (–sec ) = –1. So, it is a tangent to the conjugate hyperbola at point ( a tan , b sec ) which will obviously on the other branch.
Illustration 17: If e1 and e2 are the eccentricities of the hyperbola and its conjugate hyperbola, prove that e1-2 + e2-2 = 1.
Solution: The eccentricity e1 of the given hyperbola is obtained from
b2 = a2(e12 - 1) ….(1)
The eccentricity e2 of the conjugate hyperbola is given by
a2 = b2 (e22 -1) ….(2)
Multiply (1) and (2).
We get 1 = (e12 - 1) (e22 - 1)
0 = e12 e22 - e12 - e22 e1-2 + e2-2 = 1
ASYMPTOTE OF HYPERBOLA
The straight line, to which the tangent to a curve tends as the point of contact tends to approach infinity, is called an asymptote of the curve. In other words, an asymptotes tends to touch the curve at infinity.
Asymptote of a curve may also be defined in another way - A straight line that touches the given curve at infinity but the line itself is not at infinity is called an asymptote to the given curve.
Equation of the asymptotes:
Let y = mx + c be an asymptote of the hyperbola …..(i)
Substituting the value of y in (1),
or (a2m2 – b2) x2 + 2a2mcx + a2(b2 + c2) = 0 …..(ii)
If the line y = mx + c is an asymptote to the given hyperbola, then it touches the hyperbola at infinity. So both roots of (2) must be infinite.
a2m2 – b2 = 0 & -2a2 mc = 0 then m = & c = 0
substituting the value of m & c in y = mx + c, we get
y = .Hence the equation of the asymptotes is
Alternative Method:
Let the equation of the hyperbola be
The tangent at P(x1, y1) on it is - = 1. But (x1, y1) lies on the hyperbola
Eliminating y1 from the above equations, we find that the equations of two tangents to the curve at the point with abscissa x1 are
Taking limits when x1 tends to infinity, we have the equations of the asymptotes as .
Note:
(i) There are two asymptotes both passing through the centre and equal inclined to the axis of x the inclination being tan–1 (b/a).
(ii) The angle between two asymptotes is 2 tan–1 (b/a).
(iii) The difference between the second degree curve and pair of asymptotes is constant.
(iv) A hyperbola and its conjugate hyperbola have the same asymptotes.
(v) The equation of the hyperbola and that of its pair of asymptotes differ by a constant. For example, if S = 0 is the equation of the hyperbola, then the combined equation of the asymptotes is given by S + K = 0. The constant K is obtained from the condition that the equation S + K = 0 represents a pair of lines. Finally the equation of the corresponding conjugate hyperbola is S+ 2K = 0.
(vi) If b = a then reduces to x2 – y2 = a2. The asymptotes of rectangular hyperbola x2 – y2 = a2 are y = x which are at right angles.
Illustration 18: Find the equation of the of the hyperbola whose asymptotes are x + 2y + 3 = 0 and 3x + 4y + 5 = 0 and which passes through the point
(1, -1). Find also the equation of the conjugate hyperbola.
Solution: Combined equation of asymptotes is (x + 2y + 3) (3x + 4y + 5) = 0
or 3x2 + 10xy + 8y2 + 14x + 22y + 15 = 0 ….(i)
Also we know that the equation of the hyperbola differs from that of asymptotes by a constant.
Let the equation of the equation of the hyperbola differs from that of asymptotes by a constant.
Let the equation of the hyperbola be
3x2 + 10xy + 8y2 + 14x + 22y + = 0 ….(ii)
Since it passes through (1, -1) then
3(1)2 + 10(1)(-1) + 8(-1)2 + 14(1) + 22 (-1) + = 0
3 -10 + 8 + 14 – 22 + = 0 = 7
From (2), equation of hyperbola is
3x2 + 10xy + 8y2 +14x + 22y +7 = 0 ….(iii)
But we know that equation of conjugate hyperbola.
= 2 (Combined equation of asymptotes) – (Equation of hyperbola)
6x2 + 20xy + 16y2 + 28x + 44y + 30 – 3x2 – 10 xy – 8y2 – 14x – 22y – 7 = 0
or 3x2 + 10xy + 8y2 + 14x + 22y + 23 =0
Illustration 19: A tangent to the parabola x2 = 4ay meets the hyperbola xy = c2 in two points ‘P’ and ‘Q’. Prove that the mid point of PQ lies on a parabola.
Solution: Let ( h,k) be the mid point of PQ
Equation of PQ T = S1
kx + hy – 2hk = 0 ……(1)
any point on x2=4ay is ( t , t2/4a)
equation of tangent
tx = ……………(2)
tx – 2ay = t2/2
Comparing (1) and (2)
t = -2ak/h also
h2= -a/2 k x2 =-ay/2 Which is a parabola
Illustration 20: The equation of the line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one of its asymptotes is 3x – 4y - 6 = 0 , then find the equation of the other asymptote.
Solution: Let centre of hyperbola (a , b)
Both lines pass through centre
\ a - b -1 = 0 , 3a - 4b - 6 = 0
a = -2 , b = -3
slope of given asymptotes = 3/4
slope of required asymptotes = -4/3
\Equation is (y + 3) = ( x + 2)
4x + 3y + 17 = 0
Exercise 21: Find the hyperbola whose asymptotes are 2x - y = 3 and 3x + y - 7 = 0 and which passes through the point (1, 1).
Solution : The equation of the hyperbola differs from the equation of the asymptotes by a constant
The equation of the hyperbola with asymptotes 3x + y - 7 = 0 and 2x - y = 3 is (3x + y - 7) (2x - y - 3) + k = 0
It passes through (1, 1) k =-6
Hence the equation of the hyperbola is (2x - y - 3) (3x + y - 7) = 6.
Exercise 22: Find the equation of the hyperbola having 3x – 4y + 7 = 0 and 4x + 3y + 1 = 0 as its asymptotes and passing through the origin.
Solution: Let equation of hyperbola be (3x – 4y + 7)(4x + 3y + 1) + l = 0 it passes through (0, 0) Þ 7 ´ 1 + l = 0 Þ l = -7 Þ equation of hyperbola is (3x – 4y + 7)(4x + 3y + 1) = 7
Exercise 23: Find the locus of the point which moves such that the chord of contact of the point with respect to the hyperbola is normal to the ellipse ?
Solution: Let ( h , k) be the variable point chord of contact of to the hyperbola
….(1)
This is normal to the ellipse
ax sec - by cosec = a2 - b2 ….(2)
From (1) and (2)
(a2 - b2) =
RECTANGULAR HYPERBOLA
The equation of the rectangular hyperbola referred to its transverse and conjugate axes as coordinate axes is therefore x2 - y2 = a2.
Illustration 24: If the tangent and normal to a rectangular hyperbola cut off intercepts a1 and a2 on one axis, and b1 and b2 on the other, show that a1a2 + b1b2 = 0.
Solution: Let rectangular hyperbola be x2 – y2 = a2 and let (asec, a tan) be any point on this hyperbola. The equations of t angents and normals at this point are
x sec, a tan = a …..(i)
and xcos + ycot = 2a …(ii)
since (i) and (ii) cut intercepts a1, a2 on x-axis, then
a1 =
Equation of rectangular hyperbola having asymptotes as coordinate axes:
When the centre of any rectangular hyperbola be at the origin and its asymptotes concide with the coordinates axes, its equation is xy = c2.
Illustration 25: If the normal at the point ‘t1’ to the rectangular hyperbola xy = c2 meets it again at the point ‘t2’, prove that
Solution: Since the equation of the normal at to the hyperbola xy = c2 is
but this passes through then
t13t2 = -1
ASSIGNMENTS
1. If the latus rectum of an ellipse is equal to half the minor axis then its eccentricity is equal to
(A) 1/2 (B)
(C) 1/4 (D) 3/4
Ans (B)
Solution. Latus rectum is
Given
2. Asymptotes of the hyperbola xy = 5x + 4y are
(A) x = 4 , y = 5 (B) x = 5 , y = 4
(C) x = 4, y = 2 (d) x = 5 , Y = 2
Ans (A)
Solution. Let the pair of asymptotes be xy – 5x – 4y k 0
This should represent a pair of straight line
k 20
Hence equation is xy – 5x – 4y 20 0
Hence the equation are x 4 and y 5
3. If the equation 4x2 + ky2 = 18 represents a rectangular hyperbola , then k is equal to
(A) 4 (B) -4
(C) 3 (D) None of these
Ans (B)
Solution. Clearly for 4x2 ky2 18 to represent a rectangular hyperbola k 4
4. A tangent is drawn to the ellipse at point P such that it intersect the auxiliary circle at points A and B, if S and S’ are the focii of the ellipse, then the minimum value of S’A + SB is
(A) b (B) 2b
(C) 2a (D) none of these
Ans (B)
Solution. We know S’A.S’B b2
Applying AM GM
Clearly S’A S’B 2b
5. Equation of a rectangular hyperbola whose asymptotes are x = 3 and y = 5 and passing through (7, 8) is
(A) xy –3y + 5x +3 = 0 (B) xy + 3y + 5x +3 = 0
(C) xy –3y + 5x –3 = 0 (D) xy –3y – 5x +3 = 0
Ans (D)
Solution. Equation of hyperbola will be (x – 3)(y – 5) k 0
This passes through (7, 8)
k – 12
Hence equation is xy – 3y – 5x 3 0
6. The line x + my + n =0 will be a normal to the hyperbola , if
(A) (B)
(C) (D)
Solution. Equation of normal at any point P() is ax cos by cot a2 b2
If the line lx my n 0 is a normal, then
OBJECTIVE LEVEL – 2
7. Angie between tangents drawn from the point P to the hyperbola is equal to
(A) (B)
(C) (D)
Solution. Ans. (D)
Distance of ‘P’ from origin is 2 units, which is equal to radius of director circle of the hyperbola. Thus, required angle is .
8. A common tangent of 9x2 – 16y2 = 144 and x2 + y2 = 9 is
(A) (B)
(C) (D) None of these
Solution. Any tangent to is y mx c where, c2 16m2 – 9.
If it touch the circle x2 y2 9, then
c2 9 (1 m2) 16m2 – 9
Hence, common tangents are
i.e.
9. The equation of tangents drawn from the point (0, 4) to the hyperbola x2 – 4y2 = 36 are
(A) 5x – 6y + 24 and 5x + 6y – 24 = 0
(B) x – 4y + 16 = 0 and x + 4y – 16 = 0
(C) 2x – y + 4 = 0 and 2x + y – 4 = 0
(D) None of these
Solution. Any tangent will be in the form y mx c, where c2 36m2 – 9. If it passes through (0, 4) then 4 c 16 36m2 – 9 m .
Hence tangent are 6y 5x 24, 6y –5x 24
10. The locus of the mid point of the chords of , which pass through a fixed point P(x1, y1) is
(A) A circle (B) An ellipse
(C) A hyperbola (D) None of these
Solution. Let Q(h, k) be the mid–point of drawn chord. Equation o this chord will be T S1 i.e.
. If it’s passes through P(x1, y1) then 0. Thus, locus is
which is clearly a hyperbola.
11. The equation of the ellipse with e = , foci on y–axis, centre of the origin and passing through the point (6, 4) is
(A) x2 + 2y2 = 16 (B) 16x2 + 7y2 = 688
(C) 16x2 + 7y2 = 344 (D) none of these
Solution: (B)
b2 = a2
Then equation is it passes through (6, 4)
16x2 + 7y2 = 688 is ellipse.
12. The normal drawn to the ellipse at the extremity of the latus rectum passes through the extremity of the minor axis. Eccentricity of this ellipse is equal to
(A) (B)
(C) (D)
Solution: (A)
Equation of normal at , is
i.e., = (ay – b2)
It should pass through (0, –b)
= –ab – b2
a2 = ab + b2
= 1 – e2
e2 = e =
13. The equation of common tangents of the curves x2 + 4y2 = 8 and y2 = 4x are
(A) y – 2x – 4 = 0, y + 2x + 4 = 0
(B) y – 2x – 2 = 0, y + 2x + 2 = 0
(C) 2y – x – 4 = 0, 2y + x + 4 = 0
(D) 2y – x – 2 = 0, 2y + x + 2 = 0
Solution: (C)
Tangent in terms of ‘m’ for y2 = 4x is y = mx + and for is y = mx . For common tangents we must have = m = Thus, common tangents are y = + 2 and y = – – 2
i.e. 2y – x –4 = 0 and 2y + x + 4 = 0
14. The line y = x + p (p is parameter) cuts the ellipse at P and Q, then mid point of PQ lies on
(A) a2y + b2x (B) a2y + b2x = 0
(C) ay + bx = 0 (D) none of these
Solution: Solving the line and ellipse, we get
(a2 + b2)x2 + 2a2px + a2p2 – a2b2 = 0
and
mid point lies on y =
15. Normals drawn to the ellipse at point ‘P’ meets the coordinate axes at points A and B respectively. Locus of mid point of segment AB is
(A) 4x2a2 + 4y2b2 = (a2 – b2)2
(B) 4x2b2 + 4y2a2 = (a2 – b2)
(C) 16x2a2 + 16y2b2 = (a2 – b2)2
(D) 16x2b2 + 16y2a2 = (a2 – b2)
Solution: (A)
Let P = (acos, bsin). Equation of normal is,
A and B
It (h, k) be the mid– point of segment AB then 2h = cos, 2k = sin
4h2a2 + 4k2b2 = (a2 – b2)2
16. A common tangent to 9x2 + 16y2 = 144, y2 = x – 4 and (x – 6)2 + y2 = 4 is
(A) y = 3 (B) x = 4
(C) y = –3 (D) x = –4
Solution: (B)
Given curves are , y2 = (x – 4) and (x – 6)2 + y2 = 4. x = 4 is indeed the common tangent of these given curves.
17. The distance between the directrices of the ellipse is
(A) (B)
(C) (D) None of these
Solution: 4 = 9 (1-e2)
Distance between the directrices = Ans.
Hence (C) is the correct answer.
18. If a tangent of slope ‘m’ at a point of the ellipse = 1 passes through (2a, 0) and if ‘e’ denotes the eccentricity of the ellipse then
(A) m2 + e2 = 1 (B) 2m2 + e2 = 1
(C) 3m2 + e2 = 1 (D) none of these
Solution: Any tangent of slope m is y = mx , if it passes through (2a, 0), then 3a2m2 = b2 3m2 = .
Hence (C) is the correct answer.
19. The locus of a variable point whose distance from the point (5, 2) is times its distance from the line 2x + 3y = 5, is
(A) a circle (B) a hyperbola
(C) an ellipse (D) a parabola
Solution: As eccentricity e < 1, locus is an ellipse.
Hence (C) is the correct answer.
20. If F1 (0,0), F2 (3,4) and |PF1| +|PF2| =10, then the locus of P is
(A) An ellipse (B) A straight line
(C) A hyperbola (D) A line segment
Solution: As sum of distances is constant and greater than F1F2, locus is an ellipse.
Hence (A) is the correct answer.
21. The equation of the hyperbola in a standard form whose eccentricity is 2 and the distance between foci is 16 is
(A) x2 – y2 = 32 (B) x2 – y2 = 16
(C) x2 – y2 = 64 (D) none of these
Solution: 2ae = 16, e = , so a = 4 .
So the hyperbola (rectangular) is x2 y2 = 32
22. The latus rectum of the ellipse 5x2 + 9y2 = 45 is
(A) (B)
(C) (D)
Solution: Ellipse is . Latus rectum =
Hence (A) is the correct answer.
23. If the normal at the end of a latus rectum of an ellipse = 1passes through one extremity of the minor axis, then the eccentricity of the ellipse is given by the equation
(A) e2 + e – 1 = 0 (B) e2 + e + 1 = 0
(C) e4 + e2 + 1 = 0 (D) e4 + e2 – 1 = 0
Solution: Normal at is
It passes through (0, –b)
ab = a2 –b2 = a2e2 b = ae2 b2 = a2e4
a2(1 –e2) = a2e4 e4 + e2 –1 = 0.
Hence (D) is the correct answer.
24. If the normal at any point P of the ellipse meets the axes in G and g respectively, then |PG| : |Pg| is equal to
(A) a : b (B) a2 : b2
(C) b2 : a2 (D) b : a
Solution: Normal at P (a cos , b sin ) is ax sec –by cosec = a2 –b2
G = and g =
PG = , Pg =
Hence (C) is the correct answer.
25. The foci of the ellipse 9x2 + 4y2 –18x –24y + 9 = 0 are
(A) (1, 3 ) (B) (1, 3 2 )
(C) (1 , 3) (D) (1 2 , 3)
Solution: x = 1 is the major axis
e2 = 1 – = e =
(1, 3) is the centre of the ellipse so foci are (1, 3 ).
Hence (A) is the correct answer.
26. The equation of the chord of contact of tangents from (2, 3) to the rectangular hyperbola xy = 9 is
(A) 3x + 2y = 18 (B) 2x – 3y = 18
(C) 2x + 3y = 9 (D) 3x + 2y = 9
Solution: T = 0 xy1 + yx1 = 18 3x + 2y = 18.
27. P is a variable point on the ellipse = 1 with AA as the major axis. Then the maximum value of the area of the triangle APA is
(A) ab (B) 2ab
(C) ab/2 (D) None of these.
Solution: Area of APA will be maximum when the height of triangle will be maximum as the base is fixed. Maximum height will be equal to b .
Maximum area of APA = 2a b = ab
Hence (A) is the correct answer
28. The centre of the ellipse 25x2 + 25y2 – 14xy – 36x – 36y = 108, is
(A) (0, 0 (B) (1, 1)
(C) (1, 0) (D) (0, 1).
Solution: On simplification equation of the ellipse is
9(x + y -2 )2 + 16( x- y )2 = 144 . . . . (1)
Differentiating w. r. t. x treating y as constant, we get
18( x+ y –2 ) + 32 ( x-y ) = 0
Upon solving we have the centre of the ellipse = (1, 1)
Hence (B) is the correct Answer.
29. S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is
(A) 1/4 (B) 1/3
(C) 1/2 (D) 2/3
Solution: Let the equation of ellipse be .
S ( -ae, 0)
T ( ae , 0)
B ( 0, b ) SB =
Also SB2 = ST2 4 a2e2 = a2e2 + b2
or 3 a2e2 = a2( 1- e2)
= a2 – a2e2 e2 = 1/4 e = 1/2
Hence (C) is the correct answer.
30.. The foci of the ellipse 25 (x + 1)2 + 9 (y + 2)2 = 225 are at
(A) (-1, 2) and (-1, -6) (B) (-2, 1) and (-2, 6)
(C) (-1, -2) and (-2, -1) (D) (-1, -2) and (-1, -6).
Solution: The given equation is 25 ( x+ 1)2 + 9 ( y + 2) 2 = 225
Or,
Centre of the ellipse ( -1, -2 )
and a = 3 , b = 5, so that a< b 3 = 5
or e2 = 1- 9/ 25 = 16/25 e = 4/ 5
Hence foci are ( -1, -2 - 5 4/5) and ( -1 , -2 + 5 4/5)
i.e. foci are (-1 , -6 ) and (-1, 2) .
Hence (A) is the correct answer .
31. The locus of the point (x, y) satisfying the equation
= 4, is
(A) ellipse (B) circle
(C) straight line (D) line segment
Solution: The only possibility is M because in all the other cases sum of length will be more than 4.
Hence (D) is the correct answer.
32. The equation (10x – 5)2 + (10y – 5)2 = (3x + 4y –1)2 represents
(A) a pair of straight lines (B) an ellipse
(C) a hyperbola (D) a circle
Solution: ,
so e = , so, the equation represents an ellipse.
33. The equation of the normal to the ellipse at = is
(A) ax –by = (B) x =
(C) y = (D) x = 0
Solution: At = , the point (0, b), so the normal at that point is yaxis
34. The product of perpendicular drawn from the two foci of the ellipse to the tangent at any point of ellipse is
(A) a2 (B) 4a2
(C) 4b2 (D) b2
Solution: Product = = b2
35. The equation , r > 1 represents
(A) a circle (B) a parabola
(C) an ellipse (D) none of these.
Solution: As, r > 1, 1 r < 0 and (1 + r) < 0
so, L.H.S. = negative, but R.H.S. = positive. So that is not possible.
36. If the normal at the point P() to the ellipse intersects it again at the point Q(2), then cos is equal to
(A) 2/3 (B) 2/3
(C) 3/4 (D) none of these.
Solution: = 14 5; as it passes through ( cos 2, sin 2)
so, = 9
28 cos 10 cos = 9
18 cos2 9 cos 14 = 0
(3 cos + 2) (6 cos 7) = 0 cos =
37. The ellipse = 1 and the straight line y = mx + c intersect in real points only if
(A) a2m2 < c2 – b2 (B) a2m2 > c2 – b2
(C) a2m2 c2 – b2 (D) c b.
Solution: = 1
(b2 + a2 m2) x2 + 2 mca2 x + a2 (c2 b2) = 0
D > 0 m2c2a4 a2 (c2 b2)(b2 + a2m2) > 0 b2 + a2m2 > c2
38. Latus rectum of ellipse 4x2 + 9y2 –8x –36y + 4 = 0 is
(A) (B)
(C) (D)
Solution: 4 (x 1)2 + 9 (y 2)2 = 36 = 1
so, L.R. = = =
39. If , are the ends of a focal chord of an ellipse of eccentricity e, then tan . tan is equal to
(A) (B)
(C) (D)
Solution: Equating slope of PS and PS we get,
e cos = cos
= tan tan
40. If p, q are the segments of a focal chord of an ellipse b2x2 + a2y2 = a2b2 (a > b), then
(A) a2 (p + q) = 2bpq (B) b2 (p + q) = 2a(pq)
(C) a(p + q) = 2b2pq (D) b (p + q) = 2a2pq
Solution: H.M. of segments of any focal chord = semi latus rectum
41. Tangents to the ellipse b2x2 + a2y2 = a2 b2 are perpendicular to each other. The locus of their point of intersection is
(A) x2 + y2 = a2 + b2 (B) x2 - y2 = a2 + b2
(C) x2 + y2 = a2 - b2 (D) x2 - y2 = a2 - b2
Solution: The locus is the director circle i.e. x2 + y2 = a2 + b2
42. The equation + = 1 represents an ellipse if
(A) a < 4 (B) a > 4
(C) 4 < a < 10 (D) a > 10.
Solution: 10 a > 0 and 4 a > 0 a < 4
43. If the conics = 1 (a > b) and x2 –y2 = c2 cut orthogonally then
(A) a2 + b2 = 2c2 (B) b2 – a2 = 2c2
(C) a2b2 = 2c2 (D) a2 – b2 = 2c2
Solution: y for hyperbola = so, y for ellipse = (orthogonal cut)
b2x12 = a2y12, also, b2x12 + a2y12 = a2b2 x12 = and y12 =
also x12 y12 = c2 a2 b2 = 2c2.
44. The eccentricity of the conic represented by x2 –y2 –4x + 4y + 16 = 0 is
(A) 1 (B)
(C) 2 (D) 1/2
Solution: We have x2 –y2 –4x + 4y + 16 = 0
or, (x2 –4x) –(y2 –4y) = –16
or, (x2 –4x + 4) –(y2 –4y + 4) = –16
or (x –2)2 –(y –2)2 = –16 or,
Shifting the origin at (2, 2), we obtain , where x = X + 2, y = Y = 2
This is a rectangular hyperbola, whose eccentricity is always .
45. If m is a variable, the locus of the point of intersection of the lines and is a/ an
(A) parabola (B) ellipse
(C) hyperbola (D) none of these
Solution: The required locus is obtained by eliminating the variable m from the given equations of the lines. Then we have
This is clearly a hyperbola.
46. If the chords of contact of tangents from two points (x1, y1) and (x2, y2) to the hyperbola = 1 are at right angles, then is equal to
(A) – (B) –
(C) – (D) –
Solution: The equations of the chord of contact of tangents from (x1, y1) and (x2, y2) to the given hyperbola are
……(1)
and ……(2)
lines (1) and (2) are at right angles.
= – .
47. The equation of the chord joining two points (x1, y1) and (x2, y2) on the rectangular hyperbola xy = c2 is
(A) (B)
(C) (D)
Solution: The midpoint of the chord is .
The equation of the chord interms of its midpoint is T = S
or, x
x(y1 + y2) + y(x1 + x2) = (x1 + x2) (y1 + y2) .
48. The equation of a line passing through the centre of a rectangular hyperbola is x-y -1 = 0. If one of its asymptotes is 3x - 4y - 6 = 0, the equation of the other asymptote is
(A) 4x - 3y + 17 = 0 (B) -4x - 3y + 17 = 0
(C) -4x + 3y + 1 = 0 (D) 4x + 3y + 17 = 0
Solution: The centre is the point of intersection of the given lines
i.e. ( 2, 3) so the other asymptote passing through ( 2, 3)
and perpendicular to the first asymptote will be 4 (x + 2) + 3 (y + 3) = 0.
49. If x = 9 is the chord of contact of the hyperbola x2 –y2 = 9, then the equation of the corresponding pair of tangents is
(A) 9x2 –8y2 + 18x –9 = 0 (B) 9x2 –8y2 – 18x + 9 = 0
(C) 9x2 –8y2 – 18x –9 = 0 (D) 9x2 –8y2 + 18x + 9 = 0
Solution: x = 9 meets the hyperbola at (9, 6 ) and (9, –6 ).
The equation of tangents at these points are 3x –2 y –3 = 0
and 3x + 2 y –3 = 0.
The combined equation of these two is 9x2 –8y2 – 18x + 9 = 0.
50. The equation given by x = , y = represents
(A) ellipse (B) hyperbola
(C) parabola (D) circle
Solution: x =
y =
represents a hyperbola.
51. The normal at t of the rectangular hyperbola xy = 16 meets the x axis at
(A) (B)
(C) (D)
Solution: Equation of normal y = t2 (x 4t), it meets xaxis at,
= t2 (x 4t) x = 4t – i.e. at .
52. The foci of the ellipse and the hyperbola coincide, then the value of b2 is
(A) 1 (B) 5
(C) 7 (D) 9
Solution: For the given ellipse a2= 16. Therefore e =
e = = .
So, the foci of the ellipse are (ae, 0) i.e. .
For the hyperbola a2 = , b2= .
The eccentricity e is given by
e =
So, 16 – b2 = = 9 b2 = 7.
53. Equation of the chord of hyperbola 5x2 – 3y2 = 45, which is bisected at (5, –3) is
(A) 9x + 25y = 88 (B) 25x + 9y = 88
(C) 9x + 9y = 88 (D) none of these
Solution: T = S1 5. 5x 3 ( 3) y = 5. 52 3 (3)2 25x + 9y = 98
54. The equation of the conic with focus at (1, –1), directrix along x –y + 1 = 0 and with eccentricity is
(A) x2 –y2 =1 (B) xy = 1
(C) 2xy –4x + 4y + 1= 0 (D) 2xy + 4x – 4y – 1= 0
Solution: Let P(x, y) be any point on the conic. Then
[Using SP = e . PM]
2xy –4x + 4y + 1 = 0
55. The equation of the tangent to the hyperbola 2x2 –3y2 = 6 which is parallel to the line y = 3x + 4 is
(A) y = 3x + 5 (B) y = 3x – 5
(C) y = 3x + 5 and y = 3x – 5 (D) none of these
Solution: The equation of the tangent to is
y = mx , where m is the slope of the tangent.
Here a2 = 3, b2 = 2 and m = 3.
Therefore the equations of the tangents are y = 3x or y = 3x 5
56. The locus of the middle points of chords of hyperbola 3x2 –2y2 + 4x –6y = 0 parallel to y = 2x is
(A) 3x –4y = 4 (B) 3y –4x + 4 = 0
(C) 4x –4y = 3 (D) 3x –4y = 2
Solution: Let (h, k) be the midpoint of a chord of the hyperbola 3x2 –2y2 + 4x –6y = 0.
Then the equation of the chord is
3hx –2ky + 2(x + h) –3(y + k) = 3h2 –2k2 + 4h –6k [Using T = S]
(3h + 2)x –(2k + 3)y –3h2 + 2k2 –2h + 3k = 0
This is parallel to y = 2x, therefore
3h + 2 = 4k + 6 3h –4k = 4
Hence, locus of (h, k) is 3x –4y = 4.
57. The value of m for which y = mx + 6 is tangent to the hyperbola is
(A) (B)
(C) (D)
Solution: If y = mx + c touches , then c2 = a2m2 –b2.
Here c = 6, a2 100, b2 = 49
36 = 100m2 –49 100m2 = 85 m =
58. The point of intersection of the curves whose parametric equations are x = t2 + 1, y = 2t and x = 2s, y = 2/s, is given by
(A) (1, -3) (B) (2, 2)
(C) (-2, 4) (D) (1, 2)
Solution: x = t2 + 1 = + 1 = + 1 = + 1 = + 1.
solving we get, x2 (x 1) = 4 x = 2 so y = 2.
59. If y = mx + c is a tangent to the hyperbola x2 3y2 = 1, then
(A) c2 = m2 (B) c2 = m2 + 1
(C) c2 > m2 (D) c2 < m2
Solution: c2 = a2m2 b2 c2 = m2 c2 < m2
60. Through the positive vertex of the hyperbola = 1 a tangent is drawn, where does it meet the conjugate hyperbola
(A) at the points (±a , b) (B) at the points (0, 0)
(C) at the points (±a, b) (D) at the points (a, ±b )
Solution: Equation of tangent is x = a, solving it with = 1.
We get the points as (a, b ).
61. The condition that the line p = x cos a + y sin a becomes a tangent to = 1 is
(A) p = a cos a – b sin a (B) p2 = a2 cos a – b2 sin a
(C) p2 = a2 cos2 a + b2 sin2 a (D) p2 = a2 cos2 a – b2 sin2 a
Solution: y = ( cot ) x + p cosec . So, p2 cosec2 = a2 cot2 b2
p2 = a2 cos2 b2 sin2 .
62. Asymptotes of the hyperbola xy = 4x + 3y are
(A) x = 3, y= 4 (B) x = 4 , y = 3
(C) x = 2 , y = 6 (D) x = 6 , y = 2.
Solution: Let it be xy 4x 3y + = 0,
now (x 3)(y 4) = xy 4x 3y + 12,
so = 12 and the asymptotes are x = 3 and y = 4.
63. The rectangular hyperbola xy = 1 cuts the circle x2 + y2 = 1 at four points. The sum of abscisae of these points is
(A) 1 (B) 0
(C) 2 (D) none of these
Solution: x2 + = 4 x4 4x2 + 1 = 0
x1 + x2 + x3 + x4 = 0.
64. e1 and e2 are the eccentricities of the hyperbolas xy = c2 and x2 – y2 = c2, then =
(A) 1 (B) 4
(C) 8 (D) 6
Solution: Both the hyperbola are rectangular, so e12 + e22 = 2 + 2 = 4.
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