Parabola — Page 4 | JEE Mathematics

Q31

If the parabolas y2 = 4b(x – c) and y2 = 8ax have a common normal, then which on of the following is a valid choice for the ordered triad (a, b, c)?

A (1, 1, 3)

B (1, 1, 0)

C

\left( {{1 \over 2},2,0} \right)

D

\left( {{1 \over 2},2,3} \right)

Correct Answer

Option A

Solution

Normal to the two given curves are y = m(x – c) – 2bm – bm3, y = mx – 4am – 2am3 If they have a common normal, then (c + 2b)m + bm3 = 4am + 2am3 $\Rightarrow$ (4a – c – 2b) m = (b – 2a)m3 $\Rightarrow$ (4a – c – 2b) = (b – 2a)m2 $\Rightarrow$ m2 =

{c \over {2a - b}} - 2

>

0 By checking all options we found (A) is right option.

Note : We get that all the options are correct for m = 0 i.e., when common normal is x-axis.

But may be in question they want common normal other than x – axis, hence answer is (A).

Q32

If y = mx + 4 is a tangent to both the parabolas, y2 = 4x and x2 = 2by, then b is equal to :

A -128

B 128

C -64

D -32

Correct Answer

Option A

Solution

Given y = mx + 4 is tangent to both the parabolas. $\therefore$ Applying condition of tangent for y2 = 4x, we get

{1 \over m}

= 4 $\Rightarrow$ m =

{1 \over 4}

For x2 = 2by line y =

{x \over 4}

+ 4 is tangent $\therefore$ x2 = 2b

\left( {{x \over 4} + 4} \right)

$\Rightarrow$ x2 -

{{bx} \over 2}

- 8b = 0 For tangent to the parabola Discriminant = 0 $\Rightarrow$

{{{b^2}} \over 4}

+ 32b = 0 $\Rightarrow$ b = -128

Q33

If one end of a focal chord AB of the parabola y2 = 8x is at

A\left( {{1 \over 2}, - 2} \right)

, then the equation of the tangent to it at B is :

A 2x – y – 24 = 0

B x – 2y + 8 = 0

C x + 2y + 8 = 0

D 2x + y – 24 = 0

Correct Answer

Option B

Solution

Given parabola y2 = 8x $\therefore$ a = 2 Let one end of focal chord is A(at2 , 2at) =

\left( {{1 \over 2}, - 2} \right)

$\therefore$ 2at = -2 $\Rightarrow$ t =

- {1 \over 2}

Other end of focal chord will be B

\left( {{a \over {{t^2}}}, - {{2a} \over t}} \right)

$\equiv$ (8, 8) $\therefore$ Equation of tangent at B is 8y = 4(x + 8) $\Rightarrow$ x – 2y + 8 = 0

Q34

The area (in sq. units) of an equilateral triangle inscribed in the parabola y2 = 8x, with one of its vertices on the vertex of this parabola, is :

A

256\sqrt 3

B

64\sqrt 3

C

128\sqrt 3

D

192\sqrt 3

Correct Answer

Option D

Solution

Let A = (2t2 , 4t) B = (2t2 , -4t) (by symmetry as equilateral triangle) tan 30o =

\frac{4t}{2t^{2}}

=

\frac{2}{t}

$\Rightarrow$ t = 2

\sqrt{3}

AB = 8t = 16

\sqrt{3}

Area of equilateral =

\frac{1}{2} \times 16\sqrt{3} \times 24

=

192\sqrt 3

Q35

Let P be a point on the parabola, y2 = 12x and N be the foot of the perpendicular drawn from P on the axis of the parabola. A line is now drawn through the mid-point M of PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is

{4 \over 3}

, then :

A MQ =

{1 \over 3}

B PN = 4

C PN = 3

D MQ =

{1 \over 4}

Correct Answer

Option D

Solution

Let P = (3t2 , 6t) ; N = (3t2 , 0) M = (3t2 , 3t) Let point Q = (h, 3t) which lies on the parabola y2 = 12x $\therefore$ 9t2 = 12h $\Rightarrow$ h =

{{3{t^2}} \over 4}

Equation of NQ y - 0 =

{{3t} \over {{{3{t^2}} \over 4} - 3{t^2}}}\left( {x - 3{t^2}} \right)

$\Rightarrow$ y =

{{ - 4} \over {3t}}\left( {x - 3{t^2}} \right)

For y-intercept of NQ, x = 0 $\therefore$ y = 4t Given y-intercept of the line NQ is

{4 \over 3}

$\therefore$ 4t =

{4 \over 3}

$\Rightarrow$ t =

{1 \over 3}

$\therefore$ MQ =

{{9{t^2}} \over 4}

=

{1 \over 4}

PN = 6t = 2

Q36

Let the latus ractum of the parabola y2 = 4x be the common chord to the circles C1 and C2 each of them having radius 2

\sqrt 5

. Then, the distance between the centres of the circles C1 and C2 is :

A 8

B 12

C

8\sqrt 5

D

4\sqrt 5

Correct Answer

Option A

Solution

For parabola y2 = 4x, Length of latus rectum = 4a = 4 C1C2 = 2(C1A) C1A =

\sqrt {{{(2\sqrt 5 )}^2} - {2^2}} = 4

$\therefore$ C1C2 = 2(4) = 8

Q37

If the common tangent to the parabolas, y2 = 4x and x2 = 4y also touches the circle, x2 + y2 = c2, then c is equal to :

A

{1 \over {\sqrt 2 }}

B

{1 \over {2\sqrt 2 }}

C

{1 \over 2}

D

{1 \over 4}

Correct Answer

Option A

Solution

y = mx + {1 \over m}

(tangent at y2 = 4x) y = mx – m2 (tangent at x2 = 4y)

{1 \over m} = - {m^2}

(for common tangent) m3 = – 1 $\Rightarrow$ m = - 1 $\therefore$ Equation of tangent y = –x –1 x + y + 1 = 0 This line touches circle whose center at (0, 0), $\therefore$ apply p( Distance from center of the circle of the line ) = r ( Radius of the circle ) c =

\left| {{{0 + 0 + 1} \over {\sqrt 2 }}} \right| = {1 \over {\sqrt 2 }}

Q38

Let L1 be a tangent to the parabola y2 = 4(x + 1) and L2 be a tangent to the parabola y2 = 8(x + 2) such that L1 and L2 intersect at right angles. Then L1 and L2 meet on the straight line :

A x + 3 = 0

B x + 2y = 0

C x + 2 = 0

D 2x + 1 = 0

Correct Answer

Option A

Solution

L1 : y2 = 4(x + 1) Equation of tangent y = m(x + 1) +

{1 \over m}

...(1) L2 : y2 = 8(x + 2) Equation of tangent y = m'(x + 2) +

{2 \over {m'}}

$\Rightarrow$ y = m'x + 2

\left( {m' + {1 \over {m'}}} \right)

....(2) Since lines intersect at right angles $\therefore$ mm' = -1 $\Rightarrow$ m' =

{ - {1 \over {m}}}

Putting it in equation (2) y =

- {1 \over m}x + 2\left( { - {1 \over m} - m} \right)

$\Rightarrow$ y =

- {1 \over m}x - 2\left( {m + {1 \over m}} \right)

....(3) From equation (1) and (3) m(x + 1) +

{1 \over m}

=

- {1 \over m}x - 2\left( {m + {1 \over m}} \right)

$\Rightarrow$

\left( {m + {1 \over m}} \right)x + 3\left( {m + {1 \over m}} \right)

= 0 $\therefore$ x + 3 = 0

Q39

The centre of the circle passing through the point (0, 1) and touching the parabola y = x2 at the point (2, 4) is :

A

\left( {{6 \over 5},{{53} \over {10}}} \right)

B

\left( {{3 \over {10}},{{16} \over 5}} \right)

C

\left( {{{ - 53} \over {10}},{{16} \over 5}} \right)

D

\left( {{{ - 16} \over 5},{{53} \over {10}}} \right)

Correct Answer

Option D

Solution

Circle passes through A(0, 1) and B(2, 4). y = x2 $\Rightarrow$

{\left. {{{dy} \over {dx}}} \right|_B}

= 4 tangent at (2,4) is (y – 4) = 4(x – 2) 4x – y – 4 = 0 Equation of circle (x - 2)2 + (y–4)2 + $\lambda$ (4x–y - 4) = 0 Passing through (0,1) $\therefore$ 4 + 9 + $\lambda$ (–5) = 0 $\Rightarrow$ $\lambda$ =

{{13} \over 5}

$\therefore$ Circle is x2– 4x + 4 + y2 – 8y + 16 +

{{13} \over 5}

[4x - y - 4] = 0 $\Rightarrow$ x2 + y2 +

\left( {{{52} \over 5} - 4} \right)

x -

\left( {8 + {{13} \over 5}} \right)

y + 20 -

{{{52} \over 5}}

= 0 $\Rightarrow$ x2 + y2 +

{{{32} \over 5}x - {{53} \over 5}y}

+

{{48} \over 5}

= 0 $\therefore$ Centre is

\left( {{{ - 16} \over 5},{{53} \over {10}}} \right)

Q40

The locus of the mid-point of the line segment joining the focus of the parabola y2 = 4ax to a moving point of the parabola, is another parabola whose directrix is :

A x = 0

B x = -

{a \over 2}

C x = a

D x =

{a \over 2}

Correct Answer

Option A

Solution

Given, equation of parabola $\Rightarrow$ y2 = 4ax Focus = S(a, 0) Let any point on the parabola be P(at2, 2at). and let the mid-point of PS be M(h, k).

$\therefore$

h{{a{t^2} + a} \over 2};k = {{2at + 0} \over 2}

\Rightarrow {t^2} = {{2h - a} \over a};t = {k \over a}

\Rightarrow {t^2} = {{{k^2}} \over {{a^2}}}

Now,

{{2h - a} \over a} = {{{k^2}} \over {{a^2}}}

\Rightarrow 2h - a = {{{k^2}} \over a} \Rightarrow {k^2} = a(2h - a)

$\therefore$ Locus of (h, k) is

{y^2} = a(2x - a)

{y^2} = 2a\left( {x - {a \over 2}} \right)

$\therefore$ The directrix of this parabola is

x - {a \over 2} = - {a \over 2} \Rightarrow x = 0