 ### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2012

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

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

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

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

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

### AIEEE 2012

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.
3

### AIEEE 2011

Equation of the ellipse whose axes of coordinates and which passes through the point $$(-3,1)$$ and has eccentricity $$\sqrt {{2 \over 5}}$$ is
A
$$5{x^2} + 3{y^2} - 48 = 0$$
B
$$3{x^2} + 5{y^2} - 15 = 0$$
C
$$5{x^2} + 3{y^2} - 32 = 0$$
D
$$3{x^2} + 5{y^2} - 32 = 0$$

## Explanation

Let the ellipse be $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$

Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$

$$\Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$

So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$
4

### AIEEE 2010

If two tangents drawn from a point $$P$$ to the parabola $${y^2} = 4x$$ are at right angles, then the locus of $$P$$ is
A
$$2x+1=0$$
B
$$x=-1$$
C
$$2x-1=0$$
D
$$x=1$$

## Explanation

The locus of perpendicular tangents is directrix

i.e., $$x=-1$$

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