Conic Sections Practice Questions - DPP for JEE

 Page 2


3. (a) Let the equation of the ellipse be .
If S and S' be the foci, then SS' = 10.
But SS' = CS + CS' = 2ae, C being the centre
...(1)
Also ...(2)
Also [using (1)]
By (2), 
But a cannot be negative,
.
The equation to the ellipse is therefore
.
4. (b) Points P and Q are (a, 0) and (2a, 0)
So, the circles are
 ? x
2
 + y
2
 – ax = 0
 ? x
2
 + y
2
 – 3ax + 2a
2
 = 0
? 
Now, S be (h, k) then
?
5. (a) Equation of the ellipse is 3x
2
 + 4y
2
 = 12
?  +  = 1 .... (1)
Page 3


3. (a) Let the equation of the ellipse be .
If S and S' be the foci, then SS' = 10.
But SS' = CS + CS' = 2ae, C being the centre
...(1)
Also ...(2)
Also [using (1)]
By (2), 
But a cannot be negative,
.
The equation to the ellipse is therefore
.
4. (b) Points P and Q are (a, 0) and (2a, 0)
So, the circles are
 ? x
2
 + y
2
 – ax = 0
 ? x
2
 + y
2
 – 3ax + 2a
2
 = 0
? 
Now, S be (h, k) then
?
5. (a) Equation of the ellipse is 3x
2
 + 4y
2
 = 12
?  +  = 1 .... (1)
Eccentricity e
1
 =  = 
So, the foci of ellipse are (1, 0) and (– 1, 0)
Let the equation of the required hyperbola be
.... (2)
Given 2a = 2 sin ? ? a = sin ?
Since the ellipse (1) and the hyperbola (2) are confocal, so
the foci of hyperbola are (1, 0) and (– 1, 0) too. If the eccentricity, of
hyperbola be e
2
 then
ae
2
 = 1 ? sin ? e
2
 = 1 ? e
2
 = cosec ?
? b
2
 = a
2
 (e
2
? 2
 – 1) = sin
2
 ? (cosec
2
 ? – 1) = cos
2 
?
? Required equation of the hyperbola is
 –  = 1 ? x
2
 cosec
2
 ? – y
2
 sec
2
 ? = 1
6. (c) Given parabola is y
2
 = 4ax ...(1)
Let the coordinates of E, F and G be respectively
Since ordinates of E, F and G are in G.P.
? (2at
2
)
2
 = (2at
1
) (2at
3
) or ...(2)
The tangents at E and G are
t
1
y = ...(3)
and t
3
y
 
= 
Solving (3) and (4), we get x = at
1
t
3
 = [from (2)]
Since the x-coordinate of the point of intersection is , the point lies
on the line x =  i.e. on the ordinate of F ( , 2at
2
).
7. (d) Since the ([P + 1], [P]) lies inside the circle
x
2
 + y
2
 – 2x – 15 = 0 [But [x + n] = [x] + n, n ? N]
? [P + 1]
2
 + [P]
2
 – 2[P + 1] – 15 < 0
([P] + 1)
2
 + [P]
2
 – 2([P] + 1) – 15 < 0
Page 4


3. (a) Let the equation of the ellipse be .
If S and S' be the foci, then SS' = 10.
But SS' = CS + CS' = 2ae, C being the centre
...(1)
Also ...(2)
Also [using (1)]
By (2), 
But a cannot be negative,
.
The equation to the ellipse is therefore
.
4. (b) Points P and Q are (a, 0) and (2a, 0)
So, the circles are
 ? x
2
 + y
2
 – ax = 0
 ? x
2
 + y
2
 – 3ax + 2a
2
 = 0
? 
Now, S be (h, k) then
?
5. (a) Equation of the ellipse is 3x
2
 + 4y
2
 = 12
?  +  = 1 .... (1)
Eccentricity e
1
 =  = 
So, the foci of ellipse are (1, 0) and (– 1, 0)
Let the equation of the required hyperbola be
.... (2)
Given 2a = 2 sin ? ? a = sin ?
Since the ellipse (1) and the hyperbola (2) are confocal, so
the foci of hyperbola are (1, 0) and (– 1, 0) too. If the eccentricity, of
hyperbola be e
2
 then
ae
2
 = 1 ? sin ? e
2
 = 1 ? e
2
 = cosec ?
? b
2
 = a
2
 (e
2
? 2
 – 1) = sin
2
 ? (cosec
2
 ? – 1) = cos
2 
?
? Required equation of the hyperbola is
 –  = 1 ? x
2
 cosec
2
 ? – y
2
 sec
2
 ? = 1
6. (c) Given parabola is y
2
 = 4ax ...(1)
Let the coordinates of E, F and G be respectively
Since ordinates of E, F and G are in G.P.
? (2at
2
)
2
 = (2at
1
) (2at
3
) or ...(2)
The tangents at E and G are
t
1
y = ...(3)
and t
3
y
 
= 
Solving (3) and (4), we get x = at
1
t
3
 = [from (2)]
Since the x-coordinate of the point of intersection is , the point lies
on the line x =  i.e. on the ordinate of F ( , 2at
2
).
7. (d) Since the ([P + 1], [P]) lies inside the circle
x
2
 + y
2
 – 2x – 15 = 0 [But [x + n] = [x] + n, n ? N]
? [P + 1]
2
 + [P]
2
 – 2[P + 1] – 15 < 0
([P] + 1)
2
 + [P]
2
 – 2([P] + 1) – 15 < 0
2[P]
2
 – 16 < 0,    [P]
2
 < 8 ...(1)
From the second circle
([P] + 1)
2
 + [P]
2
 – 2([P] + 1) – 7 > 0
? 2[P]
2
 – 8 > 0, [P]
2
 > 4 ...(2)
From (1) & (2), 4 < [P]
2
 < 8, which is not possible
? for no values of ‘P’ the point will be within the region.
8. (b) Let 
We know that z = x + iy
compare real and imaginary part, we get
x = 1 – t  t = 1 – x
and 
 
Which is a hyperbola.
9. (c) The given conic is 
Squaring both sides, 
or 
Squaring again, 
or  … (1)
Comparing the equation (1) with the equation
? A = a
2
 , H = –ab, B = b
2
 , G = – a, F = –b, C = 1
Page 5


3. (a) Let the equation of the ellipse be .
If S and S' be the foci, then SS' = 10.
But SS' = CS + CS' = 2ae, C being the centre
...(1)
Also ...(2)
Also [using (1)]
By (2), 
But a cannot be negative,
.
The equation to the ellipse is therefore
.
4. (b) Points P and Q are (a, 0) and (2a, 0)
So, the circles are
 ? x
2
 + y
2
 – ax = 0
 ? x
2
 + y
2
 – 3ax + 2a
2
 = 0
? 
Now, S be (h, k) then
?
5. (a) Equation of the ellipse is 3x
2
 + 4y
2
 = 12
?  +  = 1 .... (1)
Eccentricity e
1
 =  = 
So, the foci of ellipse are (1, 0) and (– 1, 0)
Let the equation of the required hyperbola be
.... (2)
Given 2a = 2 sin ? ? a = sin ?
Since the ellipse (1) and the hyperbola (2) are confocal, so
the foci of hyperbola are (1, 0) and (– 1, 0) too. If the eccentricity, of
hyperbola be e
2
 then
ae
2
 = 1 ? sin ? e
2
 = 1 ? e
2
 = cosec ?
? b
2
 = a
2
 (e
2
? 2
 – 1) = sin
2
 ? (cosec
2
 ? – 1) = cos
2 
?
? Required equation of the hyperbola is
 –  = 1 ? x
2
 cosec
2
 ? – y
2
 sec
2
 ? = 1
6. (c) Given parabola is y
2
 = 4ax ...(1)
Let the coordinates of E, F and G be respectively
Since ordinates of E, F and G are in G.P.
? (2at
2
)
2
 = (2at
1
) (2at
3
) or ...(2)
The tangents at E and G are
t
1
y = ...(3)
and t
3
y
 
= 
Solving (3) and (4), we get x = at
1
t
3
 = [from (2)]
Since the x-coordinate of the point of intersection is , the point lies
on the line x =  i.e. on the ordinate of F ( , 2at
2
).
7. (d) Since the ([P + 1], [P]) lies inside the circle
x
2
 + y
2
 – 2x – 15 = 0 [But [x + n] = [x] + n, n ? N]
? [P + 1]
2
 + [P]
2
 – 2[P + 1] – 15 < 0
([P] + 1)
2
 + [P]
2
 – 2([P] + 1) – 15 < 0
2[P]
2
 – 16 < 0,    [P]
2
 < 8 ...(1)
From the second circle
([P] + 1)
2
 + [P]
2
 – 2([P] + 1) – 7 > 0
? 2[P]
2
 – 8 > 0, [P]
2
 > 4 ...(2)
From (1) & (2), 4 < [P]
2
 < 8, which is not possible
? for no values of ‘P’ the point will be within the region.
8. (b) Let 
We know that z = x + iy
compare real and imaginary part, we get
x = 1 – t  t = 1 – x
and 
 
Which is a hyperbola.
9. (c) The given conic is 
Squaring both sides, 
or 
Squaring again, 
or  … (1)
Comparing the equation (1) with the equation
? A = a
2
 , H = –ab, B = b
2
 , G = – a, F = –b, C = 1
Then, 
 
and  
So we have and H
2
 – AB = 0. Hence the given equation represent a
parabola.
10. (a) Let the required point be (x
1
, y
1
). The given line
3x + 2y + 1 = 0 ...(1)
is chord of contact of the point so it must be same as the line
T = 0, i.e. 4xx
1
 – yy
1
 = 4a
2
...(2)
Comparing the coefficients of (1) and (2), we get
11. (a) Let P(h, k) be a point on the circle
Then the lengths of the tangents from P(h, k) to
 and
 are
and 
or  
(Since (h, k) lies on