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1

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
Given : A circle, $$2{x^2} + 2{y^2} = 5$$ and a parabola, $${y^2} = 4\sqrt 5 x$$.
Statement-1 : An equation of a common tangent to these curves is $$y = x + \sqrt 5 $$.

Statement-2 : If the line, $$y = mx + {{\sqrt 5 } \over m}\left( {m \ne 0} \right)$$ is their common tangent, then $$m$$ satiesfies $${m^4} - 3{m^2} + 2 = 0$$.

A
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
B
Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
C
Statement-1 is true; Statement-2 is false.
D
Statement-1 is false Statement-2 is true.

Explanation

Let common tangent be

$$y = mx + {{\sqrt 5 } \over m}$$

Since, perpendicular distance from center of the circle to

the common tangent is equal to radius of the circle,

therefore $${{{{\sqrt 5 } \over m}} \over {\sqrt {1 + {m^2}} }} = \sqrt {{5 \over 2}} $$

On squaring both the side, we get

$${m^2}\left( {1 + {m^2}} \right) = 2$$

$$ \Rightarrow {m^4} + {m^2} - 2 = 0$$

$$ \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$$

$$ \Rightarrow m = \pm 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ ( as $$m \ne \pm \sqrt 2 $$ )

$$y = \pm \left( {x + \sqrt 5 } \right),$$ both statements are correct as $$m = \pm 1$$

satisfies the given equation of statement - $$2.$$
2

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
The equation of the circle passing through the foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$, and having centre at $$(0,3)$$ is
A
$${x^2} + {y^2} - 6y - 7 = 0$$
B
$${x^2} + {y^2} - 6y + 7 = 0$$
C
$${x^2} + {y^2} - 6y - 5 = 0$$
D
$${x^2} + {y^2} - 6y + 5 = 0$$

Explanation



From the given equation of ellipse, we have

$$a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}} $$

$$ \Rightarrow e = {{\sqrt 7 } \over 4}$$

Now, radius of this circle $$ = {a^2} = 16$$

$$ \Rightarrow Focii = \left( { \pm \sqrt 7 ,0} \right)$$

Now equation of circle is

$${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16$$

$${x^2} + y{}^2 - 6y - 7 = 0$$
3

AIEEE 2012

MCQ (Single Correct Answer)
An ellipse is drawn by taking a diameter of thec circle $${\left( {x - 1} \right)^2} + {y^2} = 1$$ as its semi-minor axis and a diameter of the circle $${x^2} + {\left( {y - 2} \right)^2} = 4$$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is:
A
$$4{x^2} + {y^2} = 4$$
B
$${x^2} + 4{y^2} = 8$$
C
$$4{x^2} + {y^2} = 8$$
D
$${x^2} + 4{y^2} = 16$$

Explanation

Equation of circle is $${\left( {x - 1} \right)^2} + {y^2} = 1$$

$$ \Rightarrow $$ radius $$=1$$ and diameter $$=2$$

$$\therefore$$ Length of semi-minor axis is $$2.$$

Equation of circle is $${x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}$$

$$ \Rightarrow $$ radius $$=2$$ and diameter $$=4$$

$$\therefore$$ Length of semi major axis is $$4$$

We know, equation of ellipse is given by

$${{{x^2}} \over {\left( {Major\,\,\,axi{s^{\,\,2}}} \right)}} + {{{y^2}} \over {\left( {Minor\,\,\,axi{s^{\,\,2}}} \right)}} = 1$$

$$ \Rightarrow {{{x^2}} \over {{{\left( 4 \right)}^2}}} + {{{y^2}} \over {{{\left( 2 \right)}^2}}} = 1$$

$$ \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 4} = 1$$

$$ \Rightarrow {x^2} + 4{y^2} = 16$$
4

AIEEE 2012

MCQ (Single Correct Answer)
STATEMENT-1 : An equation of a common tangent to the parabola $${y^2} = 16\sqrt 3 x$$ and the ellipse $$2{x^2} + {y^2} = 4$$ is $$y = 2x + 2\sqrt 3 $$

STATEMENT-2 :If line $$y = mx + {{4\sqrt 3 } \over m},\left( {m \ne 0} \right)$$ is a common tangent to the parabola $${y^2} = 16\sqrt {3x} $$and the ellipse $$2{x^2} + {y^2} = 4$$, then $$m$$ satisfies $${m^4} + 2{m^2} = 24$$

A
Statement-1 is false, Statement-2 is true.
B
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
C
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
D
Statement-1 is true, Statement-2 is false.

Explanation

Given equation of ellipse is $$2{x^2} + {y^2} = 4$$

$$ \Rightarrow {{2{x^2}} \over 4} + {{{y^2}} \over 4} = 1 \Rightarrow {{{x_2}} \over 2} + {{{y^2}} \over 4} = 1$$

Equation of tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$ is

$$y = mx \pm \sqrt {2{m^2} + 4} \,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

( as equation of tangent to the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

is $$y=mx+c$$ where $$c = \pm \sqrt {{a^2}{m^2} + {b^2}} $$ )

Now, Equation of tangent to the parabola

$${y^2} = 16\sqrt 3 x$$ is $$y = mx + {{4\sqrt 3 } \over m}\,\,\,\,\,\,\,\,...\left( 2 \right)$$

( as equation of tangent to the parabola

$${y^2} = 4ax$$ is $$y = mx + {a \over m}$$ )

On comparing $$(1)$$ and $$(2),$$ we get

$${{4\sqrt 3 } \over m} = \pm \sqrt {2{m^2} + 4} $$

Squaring on both the sides, we get

$$16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2}$$

$$ \Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \Rightarrow 2{m^4} + 4{m^2} - 48 = 0$$

$$ \Rightarrow {m^4} + 2{m^2} - 24 = 0 \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0$$

$$ \Rightarrow {m^2} = 4$$ ( as $${m^2} \ne - 6$$ ) $$ \Rightarrow m = \pm 2$$

$$ \Rightarrow $$ Equation of common tangents are $$y = \pm 2x \pm 2\sqrt 3 $$

Thus, statement - $$1$$ is true.

Statement - $$2$$ is obviously true.

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