1
MCQ (Single Correct Answer)

### JEE Main 2018 (Online) 15th April Evening Slot

A normal to the hyperbola, 4x2 $-$ 9y2 = 36 meets the co-ordinate axes $x$ and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the ocus of P is :
A
4x2 + 9y2 = 121
B
9x2 + 4y2 = 169
C
4x2 $-$ 9y2 = 121
D
9x2 $-$ 4y2 = 169

## Explanation

Given, 4x2 $-$ 9y2 = 36

After differentiating w.r.t.x, we get

4.2x $-$ 9.2.y.${{dy} \over {dx}}$ = 0

$\Rightarrow$ Slope of tangent = ${{dy} \over {dx}}$ = ${{4x} \over {9y}}$

So, slope of normal = ${{ - 9y} \over {4x}}$

Now, equation of normal at point (x0, y0) is given by

y $-$ y0 = ${{ - 9{y_0}} \over {4{x_0}}}$ (x $-$ x0)

As normal intersects X axis at A, Then

A = $\left( {{{13{x_0}} \over 9},0} \right)$ and B $\equiv$ $\left( {0,{{13{y_0}} \over 4}} \right)$

As OABP is parallelogram

$\therefore$ midpoint of OB $\equiv$ $\left( {0,{{13{y_0}} \over 8}} \right)$ $\equiv$ Midpoint of AP

So, P(x, y) $\equiv$ $\left( {{{ - 13{x_0}} \over 9},{{13{y_0}} \over 4}} \right)$     ...(i)

$\because$ (x0, y0) lies on hyperbola, therefore

4(x0)2 $-$ 9(y0)2 = 36

From equation (i) : x0 = ${{ - 9x} \over {13}}$ and y0 = ${{4y} \over {13}}$

From equation (ii), we get

9x2 $-$ 4y2 = 169

Hence, locus of point P is : 9x2 $-$ 4y2 = 169
2
MCQ (Single Correct Answer)

### 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${\sqrt 2 }$
3
MCQ (Single Correct Answer)

### 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 = ${1 \over 3}$
4
MCQ (Single Correct Answer)

### 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

### EXAM MAP

#### Joint Entrance Examination

JEE Advanced JEE Main

#### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

NEET