1

### JEE Main 2018 (Online) 16th April Morning Slot

The locus of the point of intersection of the lines, $\sqrt 2 x - y + 4\sqrt 2 k = 0$ and $\sqrt 2 k\,x + k\,y - 4\sqrt 2 = 0$ (k is any non-zero real parameter), is :
A
an ellipse whose eccentricity is ${1 \over {\sqrt 3 }}.$
B
an ellipse with length of its major axis $8\sqrt 2 .$
C
a hyperbola whose eccentricity is $\sqrt 3 .$
D
a hyperbola with length of its transverse axis $8\sqrt 2 .$

## Explanation

Here, lines are :

$\sqrt 2 x$ $-$ y + 4$\sqrt 2 k$ = 0

$\Rightarrow$$\,\,\,$ $\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....$(i)

and   $\sqrt 2 kx + ky - 4\sqrt 2 = 0\,\,\,\,\,...\left( {ii} \right)$

Put the value of y from (i) in (ii) we get;

$\Rightarrow$2$\sqrt 2$kx + 4$\sqrt 2$(k2 $-$ 1) = 0

$\Rightarrow$ x = ${{2\left( {1 - {k^2}} \right)} \over k}$, y = ${{2\sqrt 2 \left( {1 + {k^2}} \right)} \over k}$

$\therefore\,\,\,$ ${\left( {{y \over {4\sqrt 2 }}} \right)^2} - {\left( {{x \over 4}} \right)^2} = 1$

$\therefore\,\,\,$ length of transverse axis

2a = 2 $\times$ 4${\sqrt 2 }$ = 8${\sqrt 2 }$

Hence, the locus is a hyperbola with length of its transverse axis equal to 8${\sqrt 2 }$
2

### JEE Main 2018 (Online) 16th April Morning Slot

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus an its nearest vertex on the major axis is ${3 \over 2}$ units, then its eccentricity is :
A
${1 \over 2}$
B
${1 \over 3}$
C
${2 \over 3}$
D
${1 \over 9}$

## Explanation

If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively,

then distance between focus and vertex,

a $-$ ae = ${3 \over 2}$ (given)

$\Rightarrow$ $\,\,\,$ a (1 $-$ e) = ${3 \over 2}$

Length of latus rectum,

${{2{b^2}} \over a} = 4$

$\Rightarrow$ $\,\,\,$ b2 = 2a

$\Rightarrow$ $\,\,\,$ a2(1 $-$ e2) = 2a   [As b2 = a2 (1 $-$ e2)]

$\Rightarrow$ $\,\,\,$ a (1 $-$ e) ( 1 + e) = 2

Putting    a (1 $-$ e) = ${3 \over 2}$

$\Rightarrow$ $\,\,\,$ ${3 \over 2}$ (1 + e) = 2

$\Rightarrow$ $\,\,\,$ 3 + 3e = 4

$\Rightarrow$ $\,\,\,$ e = ${1 \over 3}$
3

### JEE Main 2018 (Online) 16th April Morning Slot

Let P be a point on the parabola, x2 = 4y. If the distance of P from the center of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :
A
x + 4y $-$ 2 = 0
B
x $-$ y + 3 = 0
C
x + y +1 = 0
D
x + 2y = 0

## Explanation

Let P(2t, t2) be any point on the parabola.

Center of the given circle C = ($-$ g, $-$f) = ($-$3, 0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC = ${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$ = ${{{t^2} - 0} \over {2t + 3}}$

Also, slope of tangent to parabola at P = ${{dy} \over {dx}}$ = ${x \over 2}$ = t

$\therefore$ Slope of normal = ${{ - 1} \over t}$

$\therefore$ ${{{t^2} - 0} \over {2t + 3}}$ = ${{ - 1} \over t}$

$\Rightarrow$ t3 + 2t + 3 = 0

$\Rightarrow$ (t+1) (t2 $-$ t + 3) = 0

$\therefore\,\,\,$ Real roots of above equation is

t = $-$ 1

Coordinate of P = (2t, t2) = ($-$2, 1)

Slope of tangent to parabola at P = t = $-$ 1

Therefore, equation of tangent is :

(y $-$ 1) = ($-$ 1) (x + 2)

$\Rightarrow$ x + y + 1 = 0
4

### JEE Main 2019 (Online) 9th January Morning Slot

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)

## Explanation

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.