Mathematics
1. Normals are drawn from the point P with slopes m1. m2 =  is a part of the parabola itself, then  is equal to
(a) 1
(b) 2
(c) 3
(d) – 2
(c) x = h, y = h
(c) x = k, y = k
2. The asymptotes to the hyperbola xy = hx + ky are
(a) x = k, y = h
(b) x = h, y = k
3. Consider twp concentric circles C1 : x2 + y2 - 1 = 0 and C2 : x2 + y2 – 4 = 0. A parabola is drawn through the points, where C1
meets the x-axis and having arbitrary tangent of C2 as it’s directrix. Then, the locus of the focus of drawn parabola is
(a)
4
3
x2 – y2 = 3
(b)
x2 – y2 = 3
3
4
x2 + y2 = 3
4
(c)
3
(d)
3
4
x2 + y2 = 3
4. A movable parabola touches the x and the y-axes at (1, 0) and (0, 1). Then, the locus of the focus of the parabola is
(a) 2x2 – 2x + 2y2 – 2y + 1 = 0
(b) x2 – 2x + 2y2 – 2y + 1 = 0
(c) 2x2 - 2x + 2y2 + 2y + 2 = 0
(d) 2x2 + 2x – 2y2 – 2y – 2 = 0
5. Circles are drawn taking any two focal chords of the parabola y2 = 4ax as diameters. Then the equation of common chord is
(a) a (t32 + t42 – t12 - t22) x + a (t3 + t4 – t1 – t2) y = 0
equal to
(b) a (t32 + t42 + t12 – t22) x – 2a (t3 + t4 – t1 +
(c) a (t32 + t42 – t12 – t22) x + 2a (t3 + t4 – t1 – t2) y = 0
t2) y = 0
(d) None of the above
6. The number of real tangents that can be drawn to the ellipse 3x2 + 5y2 = 32 passing through (3, 5) is
(a) 0
(b) 1
2
(c) 2
2
2
(d) infinite
2
7. The parabola y = x and 25 [(x – 3) + (y + 2) ] = (3x – 4y – 2) are equal, if  is equal to
(a) 1
(b) 2
(c) 3
(d) 6
2
8. Set of values of m for which a chord of slope m of the circle x + y = 4 touches the parabola y2 = 4x, is
𝟐− 𝟏
(a) (- , -
𝟐− 𝟏
)(
𝟐
𝟐
, )
2
(b) (- , - 1)  (1, )
(c) (-1, 1)
(d) R
9. The curve represented by x = 2 (cos t + sin t) and y = 5 (cos t – sin t) is
(a) a circle
(b) a parabola
(c) an ellipse
10. The area of the triangle inscribed in an ellipse is 2ab sin
(d) a hyperbola
−
−
−
𝟐
𝟐
𝟐
. Sin
. Sin
, where ,  and  are the eccentric angles of
the vertices, then the condition that the area of a triangle inscribed in an ellipse may the maximum, is
(a) ,  -
𝟐
𝟑
,+
𝟒
(b) ,  +
𝟑
𝟐
𝟑
,+
𝟒

𝟐
𝟑
𝟑
(c) 2,  + ,  +
𝟑

𝟐
𝟑
𝟑
(d) 2,  - ,  +
11. P is a point. Two tangents are drawn from it to the parabola y2 = 4x such that the slope of one triangle is three times the
slope of the other. The locus of P is
(a) a straight line
(b) a circle
(c) a parabola
(d) an ellipse
12. Consider the set of hyperbolas xy = k, x R. Let e1 be the eccentrically when k = 4 and e2 be the eccentricity when k = 9,
then e1 – e2 is equal to
(a) 0
(b) 1
(c) 2
(d) 3
13. The equation to the ellipse whose focus is the point (-1, 1) whose directrix is the straight line x – y + 3 = 0 and whose
eccentricity is
𝟏
𝟐
is
(a) 7x2 + 2xy + 7y2 + 10 x – 10y + 7 = 0
(b) x2 + 2xy + 10x – 10y + 3 = 0
(c) 3x2 + xy + 10x – 10y + 3 = 0
(d) None of the above
14. The area of a triangle inscribed in an ellipse bears a constant ratio to the area of the triangle formed by joining points on
the auxiliary circle corresponding to the vertices of the first triangle. This ratio is
(b) a2/b2
(a) b/a
(c) 2a/b
2
(d) None
2
15. The eccentricity of the conic represented by x – y – 4x + 4y + 16 = 0 is
(a) 1
(b) 𝟐
(c) 𝟑
(d) 1/2
2
2
16. If e and e’ be the eccentricities of two conics S = 0 and S’ = 0 and if e + e’ = 3, then both S and S’ can be
(a) hyperbolas
(b) ellipses
(c) parabolas
17. The tangent at point P (a cos , b sin ) of an ellipse
𝐱𝟐
𝐚𝟐
+
𝐲𝟐
𝐛𝟐
(d) None of these
= 1, meets its auxiliary circle in two points, the chord joining
these two points subtends a right angle at the centre, then the eccentricity of the ellipse is
(a) (1 + sin2)-1
(b) (1 + sin2)-1/2
(c) (1 + sin2)-3/2
(d) (1 + sin2)-2
18. If  and  are the eccentricity angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse
is
(a)
𝐜𝐨𝐬  +𝐜𝐨𝐬 
𝐜𝐨𝐬( + )
(b)
𝐬𝐢𝐧  −𝐬𝐢𝐧 
𝐬𝐢𝐧  − 
(c)
𝐜𝐨𝐬  −𝐜𝐨𝐬 
𝐜𝐨𝐬  − 
(d)
𝐬𝐢𝐧  +𝐬𝐢𝐧 
𝐬𝐢𝐧( + )
19. The number of values of such that the straight line y = 4x + c touches the curve
(a) 0
(b) 1
𝐱𝟐
𝟒
(c) 2
+ y2 = 1, is
(d) infinite
2
2
20. Tangents are drawn from the points on the line x – y – 5 = 0 to x + 4y = 4, then all the chords of contact passes through a
fixed point, whose coordinates are
𝟒
𝟏
𝟓
𝟓
𝟒 𝟏
(a) ( , - )
𝟒 𝟏
(b) ( , )
(c) (- , )
𝟓 𝟓
(d) None
𝟓 𝟓
21. The locus of all such points, so that sum of its distances from (-1, 2) and (-1, 7) is always 13, is
(a)
(𝐱+𝟏)𝟐
𝟑𝟔
+
(𝟐𝐲−𝟗)𝟐
𝟏𝟔𝟗
=1
(b)
𝟒(𝐱+𝟏)𝟐
𝟏𝟔𝟗
+
𝟗
𝟐
(𝐲− )𝟐
𝟑𝟔
=1
(c)
(𝐱+𝟏)𝟐
22. The foci of the hyperbola 3 (y – 1)2 – 4 (x – 2)2 = 12 are
(a) (0, 𝟕)
(b) (- 2, 1 - 𝟕)
(a) 3/13
(c) (2, 1 - 𝟕)
2
23. The eccentricity of the conic
𝟑𝟔
+
𝟗
𝟐
(𝐲− )𝟐
=1
𝟏𝟔𝟗
(d) None of these
(d) (0, - 𝟕)
2
4 (2y – x – 3) – 9 (2x + y – 1) = 80 s
(b) 𝟏𝟑/3
(c) 𝟏𝟑
(d) 3
24. The eccentricity of the rectangular hyperbola is
(a) 2
(b) 2
(c) 0
(d) None
2
25. The points of intersection of the curves whose parametric equations are x = t + 1, y = 2t and x = 2s, y =
(a) (1, - 3)
(b) (2, 2)
(c) (-2, 4)
2
𝑠
is given by
(d) (1, 2)
26. The eccentricity of the conjugate hyperbola of the hyperbola x2 – 3y2 = 1 is
(a) 2
(b) 2/ 𝟑
(c) 4
(d) 4/3
27. The equation of the tangent of the hyperbola 4x2 – 9y2 = 1 which is parallel to the line 4y = 5x + 27 is 24y – 30x = k, where
k is equal to
(a)  𝟏𝟔𝟏
(b)  𝟏𝟔𝟎
(c)  𝟏𝟕𝟎
(d) None of these
28. Let y = f (x) be a parabola, having its axis parallel to y-axis which I touched by the line y = x at x = 1, then
(a) f’ (0) = f’ (1)
(b) 2f (0) = 1 – f’ (0)
(c) f’ (1) = - 1
𝐱𝟐
29. The equation of a tangent parallel to y = x drawn to
(a) x – y + 1 = 0
(b) x – y + 2 = 0
30. The angle between the asymptotes of
-1
(a) 2 tan (b/a)
-1
𝐱𝟐
𝐚𝟐
-
𝐲𝟐
𝐛𝟐
(b) tan (b/a)
𝟑
-
𝐲𝟐
𝟐
(c) x + y – 1 = 0
(d) f (0) + f’ (0) + f’’ (0) = 1
= 1 is
(d) x + y – 2 = 0
= 1 is equal to
(c) 2tan-1 (a/b)
(d) tan-1 (a/b)
1. (b) Let the point P be (h, k),
3
𝐤
k = mh – 2m – m3, 


(1 – k/) -
(2 – h) + k = 0,

y2 = 2x – 2 2 + 3,

2
m3 + m (2 – h) + k = 0
2
k =h–2 +,

(i) represents a pair of straight line.
2
2

=
𝟒
2 = 4, and
…(i),
(- m1) (- m2) – 2,  = 2
where,  is a constant and Eq.
Here, A = 0, B = 0, C = , 2H = 1, 2G = - h and 2f = - k.
2
Then, ABC + 2FGH – AF – BG – CH = 0,
𝟒
Thus, the locus of (h, k) is
t1t2 = 2, 

2. (a) The equation of asymptotes is xy – hx – ky +  = 0

 m1m2m3 = - k  m3 = - k/
Aliter: Since, locus of P is a part of the parabola.
Normals at any two points t1 and t2 meet at P.

3
On comparing it with = 4x, we get,
- 2 2 + 2   = 2
𝐡𝐤
2
  = hk
𝐤
𝐡
𝟏
𝟏
𝟐
𝟐
𝟐
𝟒
0 + 2 (- ) (- ) ( ) – 0 – 0 - .

= 0,
On putting  = hk in Eq. (i), we get,
(x – k) (y – h) = 0
xy – hx – ky + hk = 0,
So, the asymptotes are x = k and y = h.
3. (d) Clearly, the parabola should pass through (1, 0) and (-1, 0).
Let directrix of the parabola be x cos  + y sin  = 2. If S (h, k) be the focus of this
parabola, then distance of ( 1, 0) from S and from the directrix should be same.
(h – 1)2 + k2 = (cos  - 2)2
…(i),
(h + 1)2 + k2 = (cos  + 2)2
and
On substracting Eq. (i) From Eq. (ii), we get
4h = 8 cos   cos  = h/2
2(h2 + k2 + 1) = 2 (cos2  + 4)
On adding Eqs. (i) and (ii), we get,

…(ii)
h2 + k2 + 1 = 4 + (h2/4),
𝟑

𝟒
h2 + k2 = 3,
Hence, locus of focus is
𝟑
𝟒
x2 + y2 = 3.
4. (a) Since, the x-axis and the y-axis are two perpendicular tangents to the parabola and both meet at the origin,
the directrix passes through the origin.
Let y = mx be the directrix and (h, k) be the focus.
FA = AM,

(𝐡 − 𝟏)𝟐 + 𝐤 𝟐 =
and FB = BN,

𝐡𝟐 + (𝐤 − 𝟏)𝟐 =
…(i),
𝐦
…(ii)
𝟏+𝐦𝟐
(h – 1)2 + h2 + k2 + (k – 1)2 = 1
From Eqs. (i) and (ii), we get,

𝐦
𝟏+𝐦𝟐
2x2 – 2x + 2y2 – 2y + 1 = 0,
or
is the required locus.
5. (c) Let PQ and RS be the two focal chord of the parabola and let P, Q, R and S be the points t 1, t2, t3, t4
respectively, then t1t2 = - 1 and t3t4 = - 1.
(x –
at12)
Clearly, on PQ as diameter is
2
(x – at2 ) + (y – 2at1) (y – 2at2) = 0,
x2 + y2 – a (t32 + t42) x – 2a (t3 + t4) y – 3a2 = 0
and circle on RS as diameter is
2
Equation of the common chord is
2
6. (c) Since, 3 (3) + 5 (5)
2
x2 + y2 – a (t12 + t22) x – 2a (t1 + t2) y – 3a2 = 0 …(i)

2
a (t3 + t4 –
t12
…(ii)
2
– t2 ) x + 2a (t3 + t4 – t1 – t2) y = 0.
- 32 > 0. So, the given point lies outside the ellipse. Hence,, two real tangents can be
drawn from the point to the ellipse.
7. (d) Let us recall that two parabolas are equal, if the length of their latusrectum are equal.
So, the length of the latusrectum of y2 = x is .
2
2
2
25 {(x – 3) + (y + 2) } = (3x – 4y – 2) ,

The equation of the second parabola is
(𝐱 − 𝟑)𝟐 + (𝐲 + 𝟐)𝟐 =
𝟑𝐱−𝟒𝐲−𝟐
𝟑𝟐 + 𝟒𝟐
, Clearly, it represents a
parabola having focus at (3, - 2) and equation of the directrix as 3x – 4y – 2 = 0. Length of the latusrectum
𝟑 𝐗 𝟑−𝟒 𝐗 −𝟐 − 𝟐
= 2 (distance between focus and directrix) = 2
𝟑𝟐 + (−𝟒)𝟐
= 6.Thus, the two parabolas are equal, if  = 6.
y2 = 4x is y = mx + .
𝟏
8. (a) The equation of tangent of slope m to the parabola
𝐦
This will be a chord of the circle x2 + y2 = 4, if length of the perpendicular from the centre (0, 0) is less
than the radius i.e..,

(m2 -
𝟐− 𝟏
𝟐
) > 0,
𝟏
𝐦 𝐦𝟐 + 𝟏

 4m4 + 4m2 – 1 > 0,
< 2,
(m2 -
𝟐− 𝟏
𝟐
) (m +
𝟐− 𝟏
𝟐
) > 0,


m  (- , -
(m2 𝟐− 𝟏
𝟐
) (m2 +
𝟐+ 𝟏
𝟐− 𝟏
, )
𝟐− 𝟏
𝟐
)(
𝟐
𝟐
)>0
9. (c) Here,
𝐱
𝟐
= cost + sin t and
𝐲
𝟓
𝐱
𝐲
𝐱𝟐
𝟐
𝟓
𝟖
= cos t – sin t, ( )2 + ( )2 = 2 (sin2 t + cos2 t) = 2, 
+
𝐲𝟐
= 1, which is an ellipse.
𝟓𝟎
10. (b) Let P, Q and R be the angular points of the triangle and let ,  and  be their eccentric angles respectively.
The vertices of the PQR are (a cos , b sin ),
Therefore, the area of the  PQR,
= 2ab sin
𝟐
sin
−
𝟐
sin
−
𝟐
,
Then, area of pqr = 2a2 sin
points on the auxiliary circle.
𝐚𝐫 (𝐏𝐐𝐑)
On putting b = a, we get,
(a cos , b sin ) and (a cos , b cos ).
−
𝐚𝐫 (𝐩𝐪𝐫)
𝐛
= ,
ar (PQR) =

𝐚
Let p, q and r be the corresponding
−
𝟐
𝐛
𝐚
sin
−
𝟐
sin
−
𝟐
ar (pqr)
Hence, PQR will be
maximum when pqr is maximum but pqr is maximum when it is an equilateral triangle and in that case,
 -  =  -  =  -  = 2/3.
Hence, when a triangle inscribed in an ellipse is maximum, the eccentric
angles of its angular points are ,  +
𝟐
𝟑
and  +
m2 - m + 1 = 0,
Or
3.

( )2
𝟒
=
𝐦
It passes through (, ). So,
Its roots are m1 and 3m1.
So,
 = m +
𝟏
𝐦
m1 + 3m1 = (/) and m1. m1 = 1/.
2
 3 = 16,

.
y = mx + .
2
𝟏
𝟑
𝐚
11. (c) Let P = (, ). Any tangent to the parabola is
(here, a = 1),
𝟒
Thus, the locus is 3y = 16x, which is a parabola.
12. (a) We know that, the eccentricity of xy = k for all k  R is 𝟐.
e1 = 𝟐 and also e2 = 𝟐,
hence, e1 – e2 = 0
13. (a) If P (x, y) be any point on the ellipse, S be its focus, and PN be the perpendicular from P on directrix, then
𝟏 𝐱−𝐲+𝟑 2
(
),
𝟒
𝟐
(𝐱−𝐲+𝟑)𝟐
by definition of an ellipse PS2 = e2 PN2, hence,
(x + 1)2 + (y – 1)2 =
[as focus is (-1, 1) and directrix is x – y + 3 = 0],

8 (x2 + y2 + 2x – 2y + 2)
= x2 + y2 + 9 – 2xy + 6x – 6y,

7x2 + 2xy + 7y2 + 10x – 10y + 7 = 0
=
𝟖
14. (a) Let P (a cos 1, b sin 1), Q (a cos 2, b sin 2) and R (a cos 3, b sin 3) be the vertices of the triangle
inscribed in the ellipse
𝐱𝟐
𝐚𝟐
𝐲𝟐
+
𝐛𝟐
= 1. The points on the auxiliary circle corresponding to these points are P’
(a cos 1, a sin 1), Q’ (a cos 2, a sin 2) and R’ (a cos 3, a sin 3).
𝐚 𝐜𝐨𝐬 𝟏 𝐛 𝐬𝐢𝐧 𝟏 𝟏
𝐜𝐨𝐬 𝟏 𝐬𝐢𝐧 𝟏 𝟏
𝟏
𝟏
= 𝐚 𝐜𝐨𝐬 𝟐 𝐛 𝐬𝐢𝐧 𝟐 𝟏 ,
= ab 𝐜𝐨𝐬 𝟐 𝐬𝐢𝐧 𝟐 𝟏 ,
𝟐
𝟐
𝐚 𝐜𝐨𝐬 𝟑 𝐛 𝐬𝐢𝐧 𝟑 𝟏
𝐜𝐨𝐬 𝟑 𝐬𝐢𝐧 𝟑 𝟏
𝐚 𝐜𝐨𝐬 𝟏 𝐚 𝐬𝐢𝐧 𝟏 𝟏
𝐜𝐨𝐬 𝟏 𝐬𝐢𝐧 𝟏 𝟏
𝟏
𝟏
= 𝐚 𝐜𝐨𝐬 𝟐 𝐚 𝐬𝐢𝐧 𝟐 𝟏 ,
= a2 𝐜𝐨𝐬 𝟐 𝐬𝐢𝐧 𝟐 𝟏 ,
𝟐
𝟐
𝐚 𝐜𝐨𝐬 𝟑 𝐚 𝐬𝐢𝐧 𝟑 𝟏
𝐜𝐨𝐬 𝟑 𝐬𝐢𝐧 𝟑 𝟏
15. (b)
1 = Area of PQR
and
2 = Area of P’Q’R’
Clearly,
𝟏
𝟐
=
𝐛
𝐚
= constant.
16. (a)
x2 + y2 = a2
17. (b) Equation of the auxiliary circle is
Equation of the tangent at point P (a cos , b sin ) is,
…(i)
(x/a) cos  + (y/b) sin  = 1
…(ii)
which meets the auxiliary circle at points A and B. Equation of pair of lines OA and OB is obtained by
x2 + y2 = a2 ( cos  +
making homogenous Eq. (i) With the help of Eq. (ii) as
(1 – cos2) x2 -
𝟐𝐱𝐲 𝐚 𝐬𝐢𝐧  𝐜𝐨𝐬 
𝐛
+ (1 -
𝐚𝟐
sin2 ) y2 = 0,
𝐛𝟐
Coefficient of x2 + Coefficient of y2 = 0,
𝐚𝟐

sin2  (1 -

(1 + sin2 ) e2 = 1,
𝐛𝟐
) + 1 = 0,
But
e=
𝐲
𝐚
𝐛
sin )2
AOB = 900
1 – cos2  + 1 – (a2/b2) sin2 = 0

 (a2 – b2) sin2  = b2, 

𝐱
a2e2 sin2  = a2 (1 – e2)
= (1 + sin2 )-1/2
𝟏
𝟏+𝐬𝐢𝐧𝟐 
18. (d) The equation of a chord joining points having eccentric angles  and  is given by
𝐱
𝐚
cos (

+
𝟐
)+
e=
𝐲
𝐛
sin (
−
𝐜𝐨𝐬( 𝟐 )
+
𝐜𝐨𝐬(
)
𝟐
,
+
𝟐
) = cos (

−
𝟐
).
e=
If it passes thought (ae, 0), then,
+
−
𝟐 𝐬𝐢𝐧 𝟐 𝐜𝐨𝐬( 𝟐 )
+
+
𝟐 𝐬𝐢𝐧
𝐜𝐨𝐬(
)
𝟐
𝟐
,
19. (c) We know that, the line y = mx + c touches the curve,
Here, a2 = 4, b2 = 1, m = 4,
or

e=
𝐱𝟐
𝐚𝟐
+
e cos (
+
𝟐
) = cos (
𝐬𝐢𝐧 + 𝐬𝐢𝐧
𝐬𝐢𝐧( + )
𝐲𝟐
𝐛𝟐
= 1 iff c2 = a2 m2 + b2.
c2 = 64 + 1  c =  𝟔𝟓
−
𝟐
)
20. (a) Let A (x1, x1 – 5) be a point on x – y – 5 = 0, then chord of contact of x 2 + 4y2 = 4 wrt A is
xx1 + 4y (x1 – 5) = 4,
(x + 4y) x1 – (20y + 4) = 0,

X + 4y = 0 and 20y + 4 = 0,
𝟏
y=-

(or from P + Q = 0),
𝟒
and x = ,
𝟓
It is passes through a fixed point.
𝟒
𝟏
𝟓
𝟓
The coordinates of fixed point are ( , - ).
𝟓
21. (a) Since, segment joining (-1, 2) and (-1, 7) is parallel to y-axis, therefore ellipse is vertical.
2be = 5, 2b = 13,
e = 5/13,

a2 = b2 (1 – e2) = 36,
or
(𝐱+𝟏)𝟐
Therefore, equation of the required ellipse is
𝟑𝟔
+
𝟗
𝟐
𝟒 (𝐲− )𝟐
𝟏𝟔𝟗
𝟗
and centre is (- 1, ).
𝟐
= 1,
i.e..,
Now, b2 = a2 (e2 – 1)
+
𝟑𝟔
(𝟐𝐲−𝟗)𝟐
𝟏𝟔𝟗
𝐘𝟐
3Y2 – 4X2 
22. (c) Writing y – 1 = Y and x – 2 = X, the equation of the hyperbola reduces to
Here, a2 = 4 and b2 = 3,
(𝐱+𝟏)𝟐
 3 = 4 (e2 – 1)  e2 – 1 =
 e2 =
𝟑
𝟒
𝟕
Thus, with respect to the new origin, the foci are
(0  ae) i.e.., (0  2.
Hence, with respect to the old origin, the foci are
(0 + 2, 1  𝟕) i.e.., (2, 1  𝟕)
𝟐
-
𝟒
𝟕
𝐗𝟐
𝟑
= 1.
= 1,
 e=
𝟒
𝟕
𝟐
) = (0,  𝟕),
23. (b) Here, 2y – x – 3 = 0 and 2x + y – 1 = 0 are perpendicular to each other.
Therefore, the equation of the conic can be written as
4[

𝟐𝐲−𝐱−𝟑 2
]
𝟓
–9[
𝟐𝐱+𝐲−𝟏 2
]
𝟓
= 60.
On putting
𝟒
𝟗
 e2 = 1 +
𝟒
𝟏𝟑
=
𝟗
𝟗
,
or
a = b  b = a2,
25. (b) On eliminating t from x = t2 + 1, y = 2t, we get,
𝟐
On substituting x = 2s and y =
2
x=
2
= X and
𝟓
𝐗𝟐
1 = 2s – s ,

2s – s – 1 = 0,

s = 1,
or
x = 2s = 2 and y =
𝐘𝟐
-
𝟒
𝟑
( )𝟐
𝟐𝐱+𝐲−𝟏 2
𝟐𝟐 + 𝟏𝟐
𝟐𝐱+𝐲−𝟏
𝟓
] = 80
= Y.
= 1,
b2 = a2 (e2 – 1) 
𝟏𝟔
𝟗
= 4 (e2 – 1),

𝐲𝟐
𝟒
a2 (e2 – 1) = a2,
+ 1  y2 = 4x – 4
 e= 𝟐
…(i)
2
4/s = 4.2s – 4  1/s = 2s – 1
2

𝟒
2
in Eq. (i), we get,
𝐬
] –9X5[
e = 𝟏𝟑/3
2
24. (b) In case of rectangular hyperbola,
3
𝟐𝟐 + 𝟏𝟐
𝟐𝐲−𝐱−𝟑
Here, a2 = 4 and b2 = 16/9.
which represents a hyperbola
e2 – 1 =
𝟐𝐲−𝐱−𝟑 2
4X2 – 9Y2 = 16 
The given equation can be written as,

4X5[
𝟐
𝐬

(s – 1) (2s2 + s – 1) = 0,
= 1.
So that
=2
26. (a)
𝐱𝟐
27. (a) The given equation can be written as,
We know that,
𝟏/𝟒
𝐦𝟐
𝟒
𝟏
−
𝟗
24y = 30x + k,
is a tangent to the given hyperbola.
𝟓
𝟒
and
𝟒
-
𝟏
𝟗
a2 =
=
𝐤𝟐
𝟐𝟒𝟐
,

𝟐𝟓
𝟔𝟒
-
𝟏
𝟗
=

y=
𝐚𝟐
-
𝟒
𝐲𝟐
𝟏
𝟗
= 1 for all m.
𝐛𝟐
𝟓𝟕𝟔
,

This is touched by the line y = x at x = 1.
2
and x = ax + bx + c must have equal roots.
𝟏𝟔𝟏
𝟔𝟒 𝐗 𝟗
=
𝐤𝟐
𝟓𝟕𝟔

or

2c + 1 = 1,
2f (0) + f’ (0) = 1
-
𝟐
=
𝟓
𝟒
x+
𝐤
𝟐𝟒
,

k2 =
𝟏𝟔𝟏 𝐗 𝟓𝟕𝟔
𝟔𝟒 𝐗 𝟗
= 161,
k =  𝟏𝟔𝟏
2
y = ax + bx + c
…(i)
𝐝𝐱
2a + b = 1 and a + b + c = 1,
𝟑
𝐤
𝟐𝟒
𝐝𝐲
Now,
𝐲𝟐
x+
( )(1,1) = + 1 and (b – 1)2 = 4ac
2a + b = 1 and (b – 1)2 = 4ac,

𝟐𝟒
Therefore, slope of the tangent at (1, 1) = 1

𝐱𝟐
𝟑𝟎
Hence, identifying the two equations, we get
𝐤𝟐
28. (b) The general equation of a parabola having its axis parabola to y-axis is
29. (a) Let y = x + c be a tangent to
𝐱𝟐
and b2 =
𝟏
is a tangent to the given hyperbola.
But given equation of tangent is
m=
𝐲𝟐
𝟏/𝟗
y = mx + 𝐚𝟐 𝐦𝟐 − 𝐛 𝟐 is always a tangent to the hyperbola
In our case, y = mx +
𝐦𝟐
-
Also, (1, 1) lies on Eq. (i),
or
a – c = 0  a = c,
a + b + c = 1,
= 1, then,
or
a+b+c=1
[or f (0) = c and f’ (0) = b]
2
c = 3 – 2 = 1  c =  1,
So, the required tangents are y = x  1.
30. (a) Asymptotes of the given hyperbola are y =  (bx/a).
Therefore, angle between them is 2tan-1 (b/a).
,