 ### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

The eccentricity of the hyperbola whose length of the latus rectum is equal to $$8$$ and the length of its conjugate axis is equal to half of the distance between its foci, is :
A
$${2 \over {\sqrt 3 }}$$
B
$${\sqrt 3 }$$
C
$${{4 \over 3}}$$
D
$${4 \over {\sqrt 3 }}$$

## Explanation

$${{2{b^2}} \over a} = 8$$ and $$2b = {1 \over 2}\left( {2ae} \right)$$

$$\Rightarrow 4{b^2} = {a^2}{e^2}$$

$$\Rightarrow 4{a^2}\left( {{e^2} - 1} \right) = {a^2}{e^2}$$

$$\Rightarrow 3{e^2} = 4 \Rightarrow e = {2 \over {\sqrt 3 }}$$
2

### JEE Main 2015 (Offline)

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1$$, is
A
$${{27 \over 2}}$$
B
$$27$$
C
$${{27 \over 4}}$$
D
$$18$$

## Explanation

The end point of latus rectum of ellipse

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ in first quadrant is

$$\left( {ae,{{{b^2}} \over a}} \right)$$ and the tangent at this point

intersects $$x$$-axis at $$\left( {{a \over e},0} \right)$$ and

$$y$$-axis at $$(0,a).$$

The given ellipse is $${{x{}^2} \over 9} + {{{y^2}} \over 5} = 1$$

Then $${a^2} = 9,{b^2} = 5$$

$$\Rightarrow e = \sqrt {1 - {5 \over 9}} = {2 \over 3}$$

$$\therefore$$ end point of latus rectum in first quadrant is

$$L\left( {2,\,\,5/3} \right)$$

Equation of tangent at $$L$$ is $${{2x} \over 9} + {y \over 3} - 1$$

It meets $$x$$-axis at $$A(9/2, 0)$$

and $$y$$-axis at $$B(0,3)$$

$$\therefore$$ Area of $$\Delta OAB = {1 \over 2} \times {9 \over 2} \times 3 = {{27} \over 4}$$ $$= 4 \times \left( {Area\,\,\Delta OAB} \right) = 4 \times {{27} \over 4} = 27\,\,$$ sq. units.
3

### JEE Main 2015 (Offline)

Let $$O$$ be the vertex and $$Q$$ be any point on the parabola, $${{x^2} = 8y}$$. If the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, then locus of $$P$$ is :
A
$${y^2} = 2x$$
B
$${{x^2} = 2y}$$
C
$${{x^2} = y}$$
D
$${y^2} = x$$

## Explanation

Let the coordinates of Q and P be (x1, y1) and (h, k) respectively.

$$\because$$ Q lies on x2 = 8y,

$$\therefore$$ x$$_1^2$$ = 8y ....... (1)

Again, P divides OQ internally in the ratio 1 : 3.

$$\therefore$$ $$h = {{{x_1} + 0} \over 4} = {{{x_1}} \over 4}$$ or x1 = 4h and

$$k = {{{y_1} + 0} \over 4} = {{{y_1}} \over 4}$$ or y1 = 4k Now putting x1 and y1 in (1) we get,

16h2 = 32k or, h2 = 2k

$$\therefore$$ the locus of P is given by, x2 = 2y.

4

### JEE Main 2015 (Offline)

The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$
A
meets the curve again in the third quadrant.
B
meets the curve again in the fourth quadrant.
C
does not meet the curve again.
D
meets the curve again in the second quadrant.

## Explanation

Given curve is

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

Difference $$w.r.t.x,$$

$$2x + 2x{{dy} \over {dx}} + 2y - 6y{{dy} \over {dx}} = 0$$

$${\left( {{{dy} \over {dx}}} \right)_{\left( {1,1} \right)}} = 1$$

Equation of normal at $$(1,1)$$ is

$$y=2-x$$

Solving eq. $$(1)$$ and $$(2),$$ we get $$x=1,3$$

Point of intersection $$\left( {1,1} \right),\left( {3, - 1} \right)$$

Normal cuts the curve again in $$4$$th quadrant.

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