Mathematics Must Do Problems of Parabola for NDA

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Must Do Problems of Parabola for NDA

Must Do Problems For NDA

Q 2846401373

Consider the equation of
parabola `25 [(x- 2)^2 + ( y -4)^2] = (4x- 3 y + 12)^2`
The coordinates of the focus are

(A)

`(3,-2 )`

(B)

`(2,4)`

(C)

`(-1,1 )`

(D)

None of these

Solution:

The given equation of parabola can
be written as

`(x-2)^2 + ( y-4)^2 = ( (4x -3y+ 12)/( sqrt (4^2 + (-3)^2 ) ) )^2`

`:.` The coordinates of focus are `( 2, 4)`
and the equation of directrix is

`4x- 3y + 12 = 0`

Correct Answer is `=>` (B) `(2,4)`

Q 2846401373

Consider the equation of
parabola `25 [(x- 2)^2 + ( y -4)^2] = (4x- 3 y + 12)^2`
Length of latusrectum is

(A)

8/5

(B)

4/5

(C)

16/5

(D)

None of these

Solution:

The distance of the focus from the
directrix

` = ( | 4(2) -3 (4) +12 |)/( sqrt (4^2 + (-3)^2 ) ) = 8/5`

`:.` The length of latusrectum

`= 2 xx 8/5 = 16/5`

Correct Answer is `=>` (C) 16/5

Q 2846401373

Consider the equation of
parabola `25 [(x- 2)^2 + ( y -4)^2] = (4x- 3 y + 12)^2`
The equation of the axis is

(A)

4x+ 3y+ 15= 0

(B)

4x+ 3y+ 10= 0

(C)

3x + 4y- 22 = 0

(D)

None of these

Solution:

Axis of parabola is perpendicular to
the directrix.

:. Equation of line perpendicular to

`4x - 3 y + 12 = 0` is `3x + 4y + k = 0`.

Since, `3x + 4y + k = 0` passes through
`(2, 4)`.

`:. 3(2) + 4(4) + k = 0 => k =- 22`

`:.` Equation of axis is `3x + 4y- 22 = 0`

Correct Answer is `=>` (C) 3x + 4y- 22 = 0

Q 2806101078

`P : y^2 = 8x ; E : x^2/4 + y^2/15 = 1`
Equation of a tangent common to both the
parabola P and the ellipse E is

(A)

`x - 2y+8=0`

(B)

`x pm 2y + 8 =0 `

(C)

`x+ 2y - 8 = 0`

(D)

`x pm 2y - 8 = 0`

Solution:

Equation of any tangent to the
parabola `P : y^2 = 8x` is `y = mx + 2/m`

where, m is the slope of tangent.

Since, it touches` E : x^2/4 + y^2/15 = 1`

`(2/m)^2 = 4m^2 +15 => m = pm 1/2`

Equations of the tangents are

`x pm 2y + 8 = 0`

Correct Answer is `=>` (B) `x pm 2y + 8 =0 `

Q 2806101078

`P : y^2 = 8x ; E : x^2/4 + y^2/15 = 1`
Equation of the normal at the point of contact of
the common tangent, which makes an acute
angle with the p:)sitive direction of X-axis, to the
parabola P is

(A)

`2x + y = 24`

(B)

`2x + y + 24 = 0`

(C)

`2x + y = 48`

(D)

`2x + y + 48 = 0`

Solution:

when `m = 1/2, ` the slope of the

normal is -2 and equation of the normal
to the parabola is

`y = -2x- 2(2) ( -2)- 2( -2)^3`

`=> 2x + y = 24`

Correct Answer is `=>` (A) `2x + y = 24`

Q 2806101078

`P : y^2 = 8x ; E : x^2/4 + y^2/15 = 1`
Point of contact of a common tangent to P and E
on the ellipse is

(A)

`(1/2 , 15/4 )`

(B)

`( pm 1/2, 15/4 )`

(C)

`(1/2 , -15/4 )`

(D)

`(-1/2 , pm 15/4 )`

Solution:

Equation of the tangent at

`(2 cos theta , sqrt 15 sin theta)` on the ellipse E to
the ellipse is `x/2 cos theta + y/(sqrt 15) = 1`,

if it represents the tangent

`x - 2 y + 8 = 0`, then

`(cos theta)/2 = (sin theta )/(-2 sqrt 15) = (-1)/8`

`=> cos theta = -1/4 , sin theta = (sqrt 15)/4`

and the point of contact is `(-1/2 , 15/4 )`

Similarly, the point of contact of the

other tangent is `(-1/2 , -15/4 )`

Correct Answer is `=>` (D) `(-1/2 , pm 15/4 )`

Q 2885780667

In the parabola `y^2 = 4ax`, the length of chord
passing through the vertex and inclined to the

axis at an angle `(pi/4)` is

(A)

`2a sqrt 2`

(B)

`sqrt 2 a`

(C)

`2a`

(D)

`4a sqrt 2`

Solution:

Let AP be the chord of length p,
making an angle of `(pi/4)` with the axis.

Then, coordinates of P are

`P ( p cos pi/4 , p sin pi/4) ` i.e, `(p/sqrt 2 , p/sqrt 2)`

Since, this point lies on `y^2 = 4ax`,

`we have `p^2/2 =4a * p/sqrt2 => p = 4a sqrt 2`

Correct Answer is `=>` (D) `4a sqrt 2`

Q 2855780664

An equilateral triangle is inscribed in a parabola
`y^2 = 4 ax`. whose vertex is at the vertex of the
parabola. The length of each side of the triangle
is

(A)

`2a sqrt 3`

(B)

`4a sqrt 3`

(C)

`6a sqrt 3`

(D)

`8a sqrt 3`

Solution:

Let `Delta OAB` be the equilateral
triangle.

Then `angle COB = 30^o`

Let `OA =OB =AB =b`

Then, B is `B (b cos 30^o, b sin 30^o`)

i.e. `B = (sqrt 3/2 b , 1/2 b)`

`B= (sqrt 3/2 b , 1/2 b)` lies on `y^2 = 4 ax`

`:. b^2/4 = 2 sqrt 3 ab => b = 8 sqrt 3 a`

Correct Answer is `=>` (D) `8a sqrt 3`

Q 2805680568

The coordinates of a point on the parabola
`y^2 = 8x`. whose focal distance is 4, is

(A)

(2,4)

(B)

(4,2)

(C)

(-2,-4)

(D)

(4,-2)

Solution:

Given, `a + x = 4 `or `2 + x = 4`

`:. x = 2`

On putting `x = 2` in `y^2 = 8x`, we get

`y = pm 4`

Correct Answer is `=>` (A) (2,4)

Q 2875680566

If (0, 4) nnd (0, 2) are respectively, the vertex
and focus of a parabola, then its equation is

(A)

`x^2 + 8y = 32`

(B)

`y^2 + 8x = 32`

(C)

`x^2 - 8y = 32`

(D)

`y^2 -8x = 32`

Solution:

AS= 2a. Vertex (0, 4) lies on Y-axis.
The parabola is of the form

`X^2 = - 4 AY` (downward parabola) as
focus is below the vertex.

`(x-0)^2 = -4 xx 2 (y-4)`

`:. x^2 + 8y = 32`

Correct Answer is `=>` (A) `x^2 + 8y = 32`

Q 2835680562

The equation `y^2- 2x - 2y + 5 = 0` represents

(A)

circle centred at (1, 1 )

(B)

parabola with directrix at `x = 3/2`

(C)

parabola with focus at (1, 2)

(D)

parabola with directrix at `x = -1/2`

Solution:

Given equation can be rewritten as

`(y - 1)^2 = 2(x - 2)`
or `Y^2 = 4 AX`, where `Y = y- 1`

`X = x- 2, 4A = 2`

i.e. Directrix is `X = - A => x - 2 = - 1/2`

`:. x = -1/2 + 2 = 3/2`

Correct Answer is `=>` (B) parabola with directrix at `x = 3/2`

Q 2815580469

The parametric representation `(2 + t^2, 2t + 1)`
represents

(A)

a parabola with locus at (2, 1)

(B)

a parabola with vertex at (2, 1)

(C)

an ellipse with centre at (2,1)

(D)

None of the above

Solution:

Let `x = 2 + t^2 , y = 2t + 1`

Eliminating t, we get

`(y - 1)^2 = 4(x - 2)`

which is a parabola with vertex at (2, 1).

Correct Answer is `=>` (B) a parabola with vertex at (2, 1)

Q 2885580467

The two ends of latusrectum of a parabola are
the points (3, 6) and ( -5, 6), then the focus is

(A)

(1,6)

(B)

(-1,6)

(C)

(1,-6)

(D)

(-1,-6)

Solution:

Focus is the mid-point latusrcctum.
Let the two end points of latusrectum
be `L_1 (3, 6)` and `L_2 (-5, 6 )`.

`:.` Focus , `x= (3+ (-5) )/2 = -1`,

`y = (6+6)/2 = 6`

Focus= (x,y) = (-1, 6)

Correct Answer is `=>` (B) (-1,6)

Q 2875580466

The two ends of latusrectum of a parabola are
the points (3, 6) and ( -5, 6), then the focus is

(A)

(1,6)

(B)

(-1,6)

(C)

(1,-6)

(D)

(-1,-6)

Correct Answer is `=>` (B) (-1,6)

Q 2874723656

The equation of line parallel to the line
2x + 3y + 5 = 0 and sum of whose intercepts on
the axes is 15 is

(A)

2x + 3y = 15

(B)

3x-2y+ 2 = 0

(C)

3x-2y+ 8 = 0

(D)

2x+ 3y-18=0

Solution:

Equation of line parallel to

`2x + 3 y + 5 = 0` is `2x + 3 y = lambda`

`=> x/(lambda/2) + y/(lambda/3) =1`

Given, `lambda/2 + lambda/3 = 15 => lambda = 18`

`:. ` Required line is `2x + 3y = 18`

Correct Answer is `=>` (D) 2x+ 3y-18=0

Advance Practice

Q 2589378217

Equation of tangent to the parabola `y^2 = 16x` at `P(3, 6)` is
BCECE Mains 2015

(A)

`4x - 3y + 12 = 0`

(B)

`3y - 4x - 12 = 0`

(C)

`4x - 3y - 24 = 0`

(D)

`3y - x - 24 = 0`

Solution:

Equation of tangent to parabola

`y^2 = 16x` at `P(3, 6)` is

`6y = 8(x + 3)`

`=> 3y = 4(x + 3)`

`=> 3y = 4x + 12`

`=> 3y - 4x - 12 = 0`

Correct Answer is `=>` (B) `3y - 4x - 12 = 0`

Q 2539580412

The line `x + y = 6` is a normal to the parabola `y^2 = 8x` at the point
BCECE Mains 2015

(A)

`(18,- 12)`

(B)

`(4,2)`

(C)

`(2, 4)`

(D)

`(8, 8)`

Solution:

Given that, parabola `y^2 = 8 x`

and line `x + y = 6`

By comparing the both equations with standard

equations of parabola and line, we get

`a = 2, m = -1`

`:.` Required point `= (am^2 , - 2am)`

`= [2(-1)^2 , -2(2)(-1)] = (2, 4)`

Correct Answer is `=>` (C) `(2, 4)`

Q 2418880709

The mid-point of the chord ` 2x + y-4 = 0` of the parabola `y^2 = 4x` is
BCECE Stage 1 2015

(A)

`(5/2, -1)`

(B)

`( -1, 5/2)`

(C)

`(3/2, -1)`

(D)

None of these

Solution:

Let `(h, k)` be the mid-point of the chord
`2x + y - 4 = 0` of the parabola `y^2 = 4x`. Then, its
equation is
`ky- 2 (x +h)= k^2 -4h` [using ·T' = S']

`=> 2x - ky + k^2- 2h = 0` ........(i)

Eq. (i) and `2x + y- 4 = 0` represent the same line

`therefore -k = 1` and `k^2-2h = -4`

`=> k = -1 , h = 5/2`

Hence, the required mid-point is `(5/2 ,-1)`

Q 2409580418

The two parabolas `x^2 = 4y` and `y^2 = 4x` meet in two distinct points. One of these is the origin and the other is
BCECE Stage 1 2012

(A)

`(2, 2)`

(B)

`(4, - 4)`

(C)

`(4, 4)`

(D)

`(-2, 2)`

Solution:

Given, equations of parabola are `x^2= 4y` and `y^2 = 4x` ............(i)

`therefore (x^2/4)^2 = 4x`

`=> x^4-64x = 0`

`=> x = 0 , x = 4`
On putting the value of `x` in Eq. (i), we get `y = 0` and `y = 4, - 4` (`because y = - 4` does not satisfy the equation `x^2 = 4y`) Hence, points of intersection are `(0, 0)` and `(4, 4)`.

Correct Answer is `=>` (C) `(4, 4)`

Q 2429145911

In Simpson's one-third rule the curve `y = f (x)` is assumed to be a
BCECE Stage 1 2011

(A)

circle

(B)

parabola

(C)

parabola

(D)

None of these

Solution:

In Simpson's one-third rule the curve `y = f (x)` is assumed to be a parabola

Correct Answer is `=>` (B) parabola

Q 1613312240

Let S be the focus of the parabola `y^2 = 8x` and `PQ` be
the common chord of the circle `x^2 + y^2 - 2x- 4y = 0`
and the given parabola. The area of `Delta POS` is
BITSAT 2015

(A)

`4` sq units

(B)

`3` sq units

(C)

`2` sq units

(D)

`8` sq units

Solution:

The parametric equations of the parabola `y^2 = 8x` are
`x = 2t^2` and `y = 4 t.`

On putting `x = 2t^2` and `y = 4 t `

`x^2 + y^2 - 2x - 4 y = 0,` we get

`4 t^4 + 16t^2 - 4t^2- 16t = 0`

` => 4 t^4 + 12t^2 - 16t = 0`

` => 4 t (t^3 + 3 t - 4) = 0`

` => t ( t - 1) ( t^2 + t + 4) = 0`

` => t= 0,t = 1`

`[ .: t^2 + t + 4 ne 0]`

Thus, the coordinates of points of intersection of the circle
and the parabola are `O(0, 0)` and `P (2, 4).` Clearly, these
are diametrically opposite points on the circle.
The coordinates of the focus `S` of the parabola are `(2, 0)`
which lies on the circle. So, `OSP` is a right triangle.

`:.` Area of `Delta OSP = 1/2 xxOS xxSP = 1/2xx2xx4 = 4` sq units

Correct Answer is `=>` (A) `4` sq units

Q 1552191034

The ends of a line segment are `P(1, 3)` and `Q(1, 1)`. `R` is a point on the line segment `PQ` such that `PR: QR =1 : \lambda`. If `R` is the interior point of the parabola `y^2=4x`, then :
BITSAT 2013

(A)

`\lambda \in (0, 1)`

(B)

`\lambda \in \ (-1, \ frac{3}{5}\ )`

(C)

`\lambda \in \ (-\ frac{1}{2}, -\ frac{3}{5}\ )`

(D)

None of these

Solution:

`R` externally divides the line segment joining `P` and `Q` in the ratio `1 : \lambda`

`\Rightarrow R = (\ frac{1 - \lambda}{1 - \lambda}, \ frac{1 - 3\lambda}{1 - \lambda} ) = (1, \ frac{1 - 3\lambda}{1 - \lambda}\ )`

Since, `R` is an interior point of the parabola `y^2 = 4x`, it must satisfy the condition `y^2 - 4x < 0`

`\Rightarrow (\frac{1 - 3\lambda}{1 - \lambda} )^2 - 4 < 0`

`\Rightarrow 9\lambda^2 - 6\lambda + 1 - 4(\lambda^2 - 2\lambda + 1) < 0`

`\Rightarrow 5\lambda^2 + 2\lambda -3 < 0`

`\Rightarrow 5\lambda^2 - 3\lambda + 5\lambda - 3 < 0`

`\Rightarrow (5\lambda - 3)(\lambda +1) < 0`

`\Rightarrow \lambda \in (-1, \ frac{3}{5} )`

Correct Answer is `=>` (B) `\lambda \in \ (-1, \ frac{3}{5}\ )`

Q 1572291136

If `x + y = k` is normal to `y^2 =12x`, then `k =`
BITSAT 2013

(A)

`3`

(B)

`6`

(C)

`9`

(D)

None of these

Solution:

A normal to the parabola `y^2 = 4ax` has the equation `y = mx - 2am - am^3`

Comparing this equation with the given equation `x + y = k`,

we get `m = -1` and we know `a` to be `3`.

So, the equation becomes `y + x = 6 + 3 = 9`

`\Rightarrow k = 9`

Correct Answer is `=>` (C) `9`

Q 1523591441

If the focus of parabola is at `(0, -3)` and its directrix is `y=3`, then its equation is:
BITSAT 2012

(A)

`x^2=-12y`

(B)

`x^2=12y`

(C)

`y^2=-12y`

(D)

`y^2=12x`

Solution:

Given, focus `=(0,-3)` and its direction `y=3`

`P` is any point on the parabola and `S` is the focus and `M` is on the the Directrix such that `PM` is perpendicular to directrix.

Using definition of parabola

`PS =PM\Rightarrow PS^2=PM^2`

`\Rightarrow (x-0)^2+(y+3)^2=(y-3)^2`

`\Rightarrow x^2 =(y-3)^2-(y+3)^2=2y(-3-3)=-12y`

Correct Answer is `=>` (A) `x^2=-12y`

Q 1560178915

If `2x + 3y + 12 = 0` and `x - y + 4 = 0` are conjugate with respect to the parabola `y^ 2 = 8x`, then `lambda` is equal to
BITSAT 2008

(A)

(B)

(C)

(D)

-3

Solution:

Using the condition that if two lines `l_1x + m_1y + n_1 = 0` and `l_2x + m_2y + n_2 = 0` are conjugate w.r.t.
parabola `y^2 = 4ax`, then

`l_1n_2 + l_2n_1 = 2am_1m_2 …… (1)`

Given conjugate lines are `2x + 3y + 12 = 0` and `x – y + 4 = 0` and equation of parabola is `y^2 = 8x`.

Here, `l_1 = 2, m_1 = 3, n_1 = 12; l_2 = 1, m_2 = –1`,

`n_2 = 4` and `a = 2`

From Eq. (1),

`2 xx 4 lambda + 1 xx 12 = 2 xx 2 xx 3 xx (–1)`

`8 lambda = – 12 – 12 , `

` lambda = –3`

Correct Answer is `=>` (D) -3

Q 1531134922

The equation of a straight line drawn through the focus of the parabola `y ^2 =- 4x` at an angle of `120^0` to the `x`-axis is :
BITSAT 2005

(A)

`y + sqrt3 (x- 1) = 0`

(B)

`y - sqrt3 (x- 1) = 0`

(C)

`y + sqrt3 (x+ 1) = 0`

(D)

`y - sqrt3 (x+ 1) = 0`

Solution:

Equation of parabola is

`y^ 2 =- 4x`

`:.` focus is `( -1, 0)`.

The equation of line passing through `(-1, 0)` is

`y - 0 = m (x + 1)` ... (i)

Since, the line makes an angle `theta = 120^0`

`m = tan theta = tan 120°`

`=> m=- sqrt3`

On putting the value of `m` in Eq. (i), we get

`y=-sqrt3 (x+ 1)`

Correct Answer is `=>` (C) `y + sqrt3 (x+ 1) = 0`

Q 2429545411

A line passing through the point of

in intersection of `x + y = 4` and `x - y = 2` makes

an angle `tan^(-1) (3/4)` with the `X-`axis. It

intersects the parabola `y^2 = 4(x- 3)` at points

`(x_1, y_1)` and `(x_2, y_2 )`, respectively. Then,

`| x_1 - x_2 |` is equal to
WBJEE 2016

(A)

`16/9`

(B)

`32/9`

(C)

`40/9`

(D)

`80/9`

Solution:

Given equations are

` x+y=4`............(i)

and `x- y = 2`...............(ii)

From Eqs. (i) and (ii), we get

` x = 3` and `y= 1`

The line through this point making an angle `tan^(-1) 3/4` with the `X`-axis is

`(y-1) = 3/4 (x-3)` `[ :. m =3/4 ]`

`=> y = (3x)/4 -5/4 = (3x-5)/4` .............(iii)

since, this line intersects the parabola

`y^2 = 4(x - 3)` at points `(x_1, y_1)` and `(x_2 , y_2 )`, respectively.

`:. ` putting `y = (3x-5)/4` in equat1on of parabola, we get

`((3x-5)/4)^2 =4(x-3)`

`=> 9x^2 - 94x +217 = 0`

`=> x_1 +x_2 = 94/9` and `x_1 x_2 = 217/9`

`:. | x_1 - x_2 | = sqrt ((x_1 +x_2)^2 -4x_1 x_2)`

`= sqrt ((94/9)^2 -4 xx 217/9)`

`=32/9`

Correct Answer is `=>` (B) `32/9`

Q 2439745612

The locus of the mid-points of all chords of
tbe parabola `y^2 = 4ax` through its vertex is
another parabola with directrix
WBJEE 2016

(A)

`x=-a`

(B)

`x=a`

(C)

`x=0`

(D)

`x= -a/2`

Solution:

Let `P (at^2, 2at)` be a moving point on the parabola

`y^2 = 4ax` and `S(a, 0)` be its focus. Let `Q (h, k)` be the
mid-point of `SP`.

Then, `h= (at^2 +a)/2`

and `k= (2 at +0)/2`

`=> (2h)/a=t^2 +1`

and `t = k/a`

`=> (2h)/a= k^2/a^2 +1`

`=> k^2=2ah -a^2` [on eliminating]

Thus, the locus of `(h,k)` is `y^2 = 2ax- a^2`

Nbw `y^2 = 2ax - a^2`

`=> (y-0)^2 = 2a (x-a/2)`

The equation of the directrix of this parabola is

`x-a/2 = -a/2 `, i.e. `x =0`

Correct Answer is `=>` (C) `x=0`

Q 2439056812

If the parabola `x^2 = ay` makes an intercept of
length `sqrt(40)` units on the line `y - 2x = 1`, then a
is equal to
WBJEE 2016

(A)

`1`

(B)

`-2`

(C)

`-1`

(D)

`2`

Solution:

Given, equation of the parabola

`x^2 = ay`

i.e., `y =x^2/a`.............(i)

and the equation of line

`y -2x =1`

i.e., `y =2x +1`...............(ii)

Fr4m Eqs. (i) and (ii), we get

`=> x^2/a= 2x +1`

`=> x^2 =2ax +a`

`=> x^2 -2ax -a =0`................(iii)

`:. x = (2a pm sqrt (4a^2 +4a) )/2`

`= (2 (a pm sqrt (a^2 +a) ) )/2 =a pm sqrt(a^2 +a)`

Po nts of intersection of the parabola and the line are

`(a+ sqrt (a^2 +a) , 2 (a+ sqrt (a^2 +a) ) +1 )`

and `(a- sqrt (a^2 +a) , 2 (a- sqrt (a^2 +a) ) +1 )`

`:.` Distance between the points `= sqrt(40)`

`:. sqrt (40) = sqrt ( (2 sqrt (a^2+a) )^2 + (4 sqrt (a^2 +a) )^2 )`

`=> 40= 4 (a^2 +a) +16 (a^2 +a)`

`=> 2=a^2 +a => a =1,-2`

Correct Answer is `=>` (A)

Q 2512245139

If `y = 4x + 3` is parallel to a tangent to the parabola `y^2 = 12x`, then its distance from the normal parallel to the given line is
WBJEE 2014

(A)

`213/(sqrt 17)`

(B)

`219/(sqrt 17)`

(C)

`211/sqrt(17)`

(D)

`210/(Sqrt 17)`

Solution:

Given equation of parabola is

`y^2=12 x`..................(i)

On differentiating both sides w.r.t. x, we get

`2y(dy)/(dx)=12=> (dy)/(dx)=6/y`

Since, the normal to the curve is parallel to the line `y =4x + 3`.

`:.` Slope of normal curve= Slope of line

`=> -y/6=4`

`=> y=-24`

From Eq. (i), we get

`(-24)^2=12x`

`=> 24 xx 24=12 x`

`=> x=48`

`:.` Normal point on a curve is `(48, -24)`.

`:.` Distance from `(48, -24)` to the line `4x -y + 3 =0` is

`(4 xx 48 +24 +3)/(sqrt(4^2+1^2))=219/(sqrt 17)`

Correct Answer is `=>` (B) `219/(sqrt 17)`

Q 2572345236

The value of `lambda` for which the curve `(7x + 5)^2 + (7y + 3)^2 = lambda^2(4x + 3y - 24)^2` represents a parabola is
WBJEE 2014

(A)

`pm 6/5`

(B)

`pm 7/5`

(C)

`pm 1/5`

(D)

`pm 2/5`

Solution:

Given curve is

`(7x + 5)^2 + (7y + 3)^2 = lambda(4x + 3y- 24)^2`

`=> 49x^2 + 25+ 70x + 49y^2 + 9+ 42y`

`=lambda^2(16x^2 + 9y^2 + 576+ 24xy-144y-192x)`

`=> (49 -16lambda^2) x ^2 + (49 -9lambda^2) y^2 + (70 + 192lambda^2 ) x+ (42 + 144 lambda^2
) y- 24lambda^2xy + (25- 576lambda^2)= 0`

On comparing with

`ax^2 + 2hxy +by^2 + 2gx + 2fy + c = 0`, we get

`a =49 -16lambda^2, b =(49-9lambda^2), h =-12lambda^2`

Condition for parabola is `ab = h^2`

`:. (49-16lambda^2)(49-9lambda^2)=(-12lambda^2 )^2`

`=> 2401-441 lambda^2 -784 lambda^2+ 144lambda^4 = 144 lambda^4`

`=> 2401 -1225 lambda^2 = 0`

`=> lambda^2=((49)^2)/((35)^2)`

`=> lambda=pm 49/35`

`=> lambda= pm 7/5`

Correct Answer is `=>` (B) `pm 7/5`

Q 2522180931

The equation of the common tangent with positive slope to the parabola `y^2= 8sqrt 3 x` and the hyperbola `4x^2- y^2 = 4` is
WBJEE 2014

(A)

`y= sqrt 6 x+ sqrt 2`

(B)

`y= sqrt 6 x- sqrt 2`

(C)

`y= sqrt 3 x+ sqrt 2`

(D)

`y= sqrt 3 x -sqrt 2`

Solution:

Equation of tangent in slope form of parabola `y^2=8sqrt 3 x`

`y = mx + c`.................(i)

where, `c=a/m`

`:. c= (2 sqrt 3)/m`................(ii)

Also, tangent to the hyperbola

`4x^2-y=4`

or `x^2/1-y^2/4=1` is

`c^2=a^2m^2-b^2`

`c^2=1m^2-4`

`=> ((2 sqrt 3)/m)^2=m^2-4` [from Eq. (ii))

`=> 12/m^2=m^2-4`

`=> m^4 -4m^2 -12=0`

`=> m^4 -6m^2 +2m^2 -12=0`

`=>m^2(m^2 -6) + 2(m^2 -6) =0`

`=> (m^2+2)(m^2-6)=0`

`=>m^2-6=0` and `m^2+2 ne 0`

`=> m^2=6`

`=> m = pm sqrt 6`

i.e., `m=sqrt 6` as `m` is positive slope.

`:.` from Eq (i)

`y= sqrt 6 x +(2 sqrt 3)/(sqrt 6)`

`=> y=sqrt 6 x+ sqrt 2`

Correct Answer is `=>` (A) `y= sqrt 6 x+ sqrt 2`

Q 2562178935

The point on the parabola `y^2=64x` which is nearest to the line `4x + 3y + 35 = 0` has coordinates
WBJEE 2014

(A)

`(9,-24)`

(B)

`(1,81)`

(C)

`(4,-16)`

(D)

`(-9,-24)`

Solution:

Given equation of parabola is `y^2=64 x`................(i)

The point at which the tangent to the curve is parallel to the line is the nearest point on the curve.

On differentiating both sides of Eq. (i). we get

`2y(dy)/(dx) =64`

`=> (dy)/(dx)=32/y`

Also, slope of the given line is `-4/3`

`:. -4/3=32/y=> y=-24`

From Eq. (i), `( -24)^2 = 64x => x = 9`

`:.` Hence, the required point is `(9, -24)`.

Correct Answer is `=>` (A) `(9,-24)`

Q 2522634531

A line passing through the point of intersection of `x + y = 4` and `x - y = 2` makes an angle `tan^(-1) ( 3/4)` with the x-axis. It intersects the parabola `y^2 = 4 (x - 3)` at points `(x_1, y_1)` and `(x_2, y_2 )` respectively. Then, `| x_1 -x_2 |` is equal to
WBJEE 2013

(A)

`16/9`

(B)

`32/9`

(C)

`40/9`

(D)

`80/9`

Solution:

Given lines are ` x+y=4` and `x-y=2`

On solving these lines, we get `x = 3` and `y= 1`
Now, the equation of line which passes through the intersection point (3, 1) having slope

`theta = tan^(-1)(3/4)` is `(y-1) = 3/4(x-3)`

`=> 4y-4 = 3x-9`

`=> 3x-4y = 5` .............(i)

Now tor the intersection point of the line (i) with

parabola `y^2 = 4(x-3)` put `y = ((3x-5)/4)`

then we get `(3x-5)^2/(16) = 4(x-3)`

`=> 9x^2+25-30x = 64x-192`

`=> 9x^2-94x+217 = 0`

`=> x = (94pm sqrt(8836-7812))/(18) = (94 pmsqrt(1024))/(18)`

`x = (94pm32)/(18) = 126/18 ` or `62/18`

`=> x_1 = 21/3 = 7`

`x_2 = 31/9`

`therefore |x_1-x_2| = |7-31/9| = |32/9| = 32/9`

Correct Answer is `=>` (B) `32/9`

Q 2522034831

If `P` be a point on the parabola `y^2 = 4ax` with focus `F`. Let `Q` denote the foot of the perpendicular from `P` onto the directrix.
then `(tan angle PQF)/(tan anglePFQ) ` is
WBJEE 2013

(A)

`1`

(B)

`1/2`

(C)

`2`

(D)

`1/4`

Solution:

Equation of parabola is
`y^2 = 4ax` ..............(i)

Let the parametric coordinate of point `P` on the parabola is `(a, 2a)`.

Now `QF = 2sqrt2a`

`PQ = 2a` and `PF = 2a`

we observe that `QF^2 = PQ^2+PF^2`

`=> 8a^2 = 4a^2+4a^2 = 8a^2`

So, `DeltaQPF` form a right angle isoceles triangle.
In which, `anglePQF = anglePFQ`

`=> tan anglePQF = tan anglePFQ`

`=> (tan anglePQF)/(tan anglePFQ) = 1`

Correct Answer is `=>` (A) `1`

Q 2552134934

The area of the region enclosed between parabola `y^2 = x` and the line `y = mx` is `1/48` Then, the value of `m` is
WBJEE 2013

(This question may have multiple correct answers)

(A) `-2`

(B) `-1`

(D) `2`

Solution:

Equation of parabola is `y^2 = x` and line `y = mx` For intersection point of both curves put `x = y^2`,
we get

`y = my^2 => y(my-1) = 0`

`=> y = 0` or `y = 1/m`

then `x = 0` or `x = 1/m^2`

`therefore` Intersection points are `(0, 0)` and `P(1/m^2 ,1/m)`

`therefore` Required area

` = int_(0)^(1/m) |(y/m-y^2)|dy = |[y^2/(2m)-y^3/3]_0^(1/m)|`

` = |1/(2m^3)-1/(3m^3)| = |1/(6m^3)| = 1/48` (given)

` => 1/(6m^3) = pm 1/48 => m^3 = pm8`

Now if `m^3 = 8`

`=> m^3 = (2)^3 => m = 2`

`=> m^3 = (-2)^3 => m = -2`

Correct Answer is `=>` (A)

Q 2552867734

Let P and Q be the points on the parabola `y^2 = 4x` so that the line segment `PQ` subtends right angle at the vertex. If PQ intersects the axis of the parabola at R, then the distance of the vertex from R is
WBJEE 2012

(A)

`1`

(B)

`2`

(C)

`4`

(D)

`6`

Solution:

According to question, it is given that slope of

`PX` and `QX` are perpendicular to each other, i.e.,

slope of `PX xx` slope of `XQ = - 1`

` => (2t)/t^2 xx (2m)/m^2 = -1`

`=> t_m = -4`

`PQ : (x - t^2)/(t^2 - m^2) = (y - 2t)/( 2t- 2m)`

Let `PQ` meets axis of parabola i.e., x-axis at

`R (alpha, 0)`, then

` (alpha - t^2)/((t + m) (t - m) ) = ( 0 - 2t) /(2 (t - m))`

` => alpha - t^2 = -t^2 -tm = -t^2 + 4`

`=> alpha = 4 => AR = 4`

Correct Answer is `=>` (C) `4`

Q 2512623539

The coordinates of a moving point `P` are

`(2t^2 + 4, 4t + 6)`. Then, its locus will be a
WBJEE 2011

(A)

circle

(B)

straight line

(C)

parabola

(D)

Ellipse

Solution:

Let `x = 2t^2 + 4, y = 4t + 6`

`:. x=2 ((y-6)/4)^2 +4`

`=> (y-6)^2=8 (x-4)`

Hence, it is a locus of parabola.

Correct Answer is `=>` (C) parabola

Q 2523401341

For the real parameter `t`, the locus of the
complex number `z = (1-t^2) + i sqrt (1+ t^2)` in the
complex plane is
WBJEE 2011

(A)

an ellipse

(B)

a parabola

(C)

a circle

(D)

a hyperbola

Solution:

Let `z = x + iy`

Given, `z = ( 1-t^2 ) + i sqrt (1+t^2)`

`:. x + iy= (1-t^2) +i sqrt (1+t^2)`

On equating real and imaginary parts, we get

`x= 1 - t^2` and `y = sqrt (1+ t^2)`

`:. x+y^2 =1- t^2 + (1+t^2)`

`=> x+y^2 =2 => y^2 = -(x-2)`

Hence, it represents a parabola.

Correct Answer is `=>` (B) a parabola

Q 2503512448

The locus of the middle points of all chords
of the parabola `y^2 = 4ax` passing through the vertex is
WBJEE 2011

(A)

a straight line

(B)

an ellipse

(C)

a parabola

(D)

a circle

Solution:

Let `OB` be a chord and let `B(h, k)`.

Then, coordinate of `M` be `(h/2, k/2)`

But `(h, k)` lies on the parabola.

`:. k^2 = 4ah`

`=> (k/2)^2 xx 4 = 8a h/2`

Hence, locus of `(h/2, k/2)` is `y^2 =2ax`

Hence, it represents a parabola.

Correct Answer is `=>` (C) a parabola

Q 1672623536

The condition so that the line
`lx + my + n= 0` may touch the parabola
` y^2 =8x`
UPSEE 2016

(A)

`m^2 =8ln`

(B)

`m^2 =2ln`

(C)

`8m^2 =ln`

(D)

`2m^2 =ln`

Solution:

Eliminating x between the given equations

`y^2 =8(-(n+my)/l) => ly^2+8my+8n =0`
Given straight line touches the parabola if roots of the equation are same

`(8m)^2 = 4l8n => 2m^2 =ln`

Correct Answer is `=>` (D) `2m^2 =ln`

Q 2415591469

The tangent at `(1, 7)` to the curve `x^2 = y - 6`
touches the circle `x^2 + y^2 + 16x + 12y + c = 0`
at
UPSEE 2014

(A)

`(6, 7)`

(B)

`(- 6, 7)`

(C)

`(6, -7)`

(D)

`(- 6, -7)`

Solution:

The tangent at `(1 , 7)` to the parabola `x^2 = y - 6` is

`x = 1/2 (y + 7) - 6`

`=> 2x = y + 7 - 12`

`=> y = 2x + 5`

which is also a tangent to the circle

`x^2 + y^2 + 16x + 12 y + c = 0`

`:. x^2 + (2x + 5)^2 + 16x + 12 (2x + 5) + c = 0`

`=> 5x^2 + 60x + 85 + c = 0`

must have equal roots.

Let `alpha` and `beta` be the roots of the equation.

Then, `alpha + beta = -12 => alpha = -6 quad ( ∵ alpha = beta)`

`:. x = - 6` and `y = 2x + 5 = - 7`

`=>` Point of contact is `(- 6, -7)`.

Correct Answer is `=>` (D) `(- 6, -7)`

Q 2435123962

The axis of the parabola
`9y^2 - 16x -12y- 57= 0` is
UPSEE 2012

(A)

`3y = 2`

(B)

`x+ 3y = 3`

(C)

`2x = 3`

(D)

`y = 3`

Solution:

Since, `9y^2 -16x -12y- 57= 0`

`=> (y- 2/3)^2 = 16/9 (x+ 61/16)`

Put `y-2/3 =y` and `x+ 61/16 = x`

`=> y^2 = 4 (4/9) x`

Axis of this parabola is `y = 0`

`=> y -2/3 = 0 => y = 2/3`

`=> 3y = 2`

Correct Answer is `=>` (A) `3y = 2`

Q 2473478346

For the parabola `y ^2 + 8x - 12y + 20 = 0`
UPSEE 2011

(A)

vertex is `(2, 6)`

(B)

focus is` (0, 6)`

(C)

latusrectum `4`

(D)

axis `y = 6`

Solution:

Given parabola can be rewritten as

`(y - 6)^ 2 = - 8(x- 2)`

Which is of the form,

`y^ 2 =- 4AX`

where `Y=y-6,X=x-2, A=2`

`:.` Vertex `(2, 6)`

Focus `(0, 6)`

Latusrectum `= 4A = 4 xx 2 = 8`

Axis is `y = 6`

Correct Answer is `=>` (C) latusrectum `4`