Parabola — Page 11 | JEE Mathematics

Q101

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin, on the positive x-axis then which of the following points does not lie on it?

A (5, 2

\sqrt 6

)

B (6, 4

\sqrt 2

)

C (8, 6)

D (4, -4)

Correct Answer

Option C

Solution

So the equation of the parabola,

{\left( {y - 0} \right)^2} = 4.a\left( {x - 2} \right)

$\Rightarrow$ y2 = 4.2 (x $-$ 2) $\Rightarrow$ y2 = 8 (x $-$ 2) By checking each options you can see. point (8, 6) does not lie on the parabola.

Q102

Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x is :

A

2\sqrt 3

y = 12x + 1

B

\sqrt 3

y = x + 3

C

2\sqrt 3

y = -x - 12

D

\sqrt 3

y = 3x + 1

Correct Answer

Option B

Solution

We know, Equation of tangent to the parabola y2 = 4ax is, y = mx +

{a \over m}

$\therefore$ Equation of tangent to the parabola y2 = 4x is, y = mx +

{1 \over m}

$\Rightarrow$ m2x $-$ ym + 1 = 0 This tangent is also the tangent to the circle x2 + y2 $-$ 6x = 0 So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3 $\therefore$

\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3

$\Rightarrow$ (3m2 + 1)2 = 9(m4 + m2) $\Rightarrow$ 9m4 + 6m2 + 1 = 9m4 + 9m2 $\Rightarrow$ 3m2 = 1 $\Rightarrow$ m = $\pm$

{1 \over {\sqrt 3 }}

So, possible tangents are y =

{1 \over {\sqrt 3 }}

x +

\sqrt 3

$\Rightarrow$

\sqrt 3

y = x + 3 or y = $-$

{x \over {\sqrt 3 }}

$-$

\sqrt 3

$\Rightarrow$

\sqrt 3 y

= $-$ x $-$ 3

Q103

If

\theta

denotes the acute angle between the curves, y = 10 – x2 and y = 2 + x2 at a point of their intersection, the |tan

\theta

| is equal to :

A

8 \over 15

B

4 \over 9

C

7 \over 17

D

8 \over 17

Correct Answer

Option A

Solution

Angle between the curves is the acute angle between the tangents at the point of intersection. y = 10 $-$ x2 (for curve 1) and y = 2 + x2 (for curve 2) $\therefore$ 10 $-$ x2 = 2 + x2 $\Rightarrow$ 2x2 = 8 $\Rightarrow$ x2 = 4 $\Rightarrow$ x = 2, $-$ 2 $\therefore$ points of intersection (2, 6) and ( $-$ 2, 6)

{{dy} \over {dx}}

for curve 1 = $-$ 2x $\therefore$ Slope(m1) of curve 1 is = $-$ 2(2) = $-$ 4

{{dy} \over {dx}}

for curve 2 = 2x $\therefore$ slope (m2) of curve 2 = 2 $\times$ 2 = 4 $\therefore$ tan $\theta$ =

\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|

=

\left| {{{ - 4 - 4} \over {1 + \left( { - 16} \right)}}} \right|

=

{8 \over {15}}

Q104

Let A(4,

-

4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of

\Delta

ACB is maximum. Then, the area (in sq. units) of

\Delta

ACB, is :

A

31{1 \over 4}

B

30{1 \over 2}

C 32

D

31{3 \over 4}

Correct Answer

Option A

Solution

\Delta ABC = {1 \over 2}\left| \begin{array}{lll}4 & { - 4} & 1 \\ 9 & 6 & 1 \\ {{t^2}} & {2t} & 1 \end{array} \right|

D = 60 + 10t $-$ 10t2

{{d\Delta } \over {dt}} = 0 \Rightarrow t = {1 \over 2}

{{{d^2}\Delta } \over {d{t^2}}} = - 20 < 0

$\therefore$ max at

t = {1 \over 2}

max area

\Delta = 65 - {5 \over 2}

= {{125} \over 2} = 31{1 \over 4}

Q105

If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :

A

5\sqrt 5

B

{\left( {10} \right)^{2/3}}

C

5\left( {{2^{1/3}}} \right)

D 5

Correct Answer

Option D

Solution

Vertex is (a2, 0) y2 $=$ $-$ (x $-$ a2) and x $=$ 0 $\Rightarrow$ (0, $\pm$ 2a) Area of triangle is

= {1 \over 2}.

4a.(a2) = 250 $\Rightarrow$ a3 = 125 or a = 5