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### JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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 $$\Delta OAB = {1 \over 2} \times {9 \over 2} \times 3 = {{27} \over 4}$$

By symmetry area of quadrilateral

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

### JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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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