1

### JEE Main 2017 (Online) 8th April Morning Slot

The locus of the point of intersection of the straight lines,

tx $-$ 2y $-$ 3t = 0

x $-$ 2ty + 3 = 0 (t $\in$ R), is :
A
an ellipse with eccentricity ${2 \over {\sqrt 5 }}$
B
an ellipse with the length of major axis 6
C
a hyperbola with eccentricity $\sqrt 5$
D
a hyperbola with the length of conjugate axis 3

## Explanation

Here, tx $-$ 2y $-$ 3t = 0  &  x $-$ 2ty + 3 = 0

On solving, we get;

y = ${{6t} \over {2{t^2} - 2}}$ = ${{3t} \over {{t^2} - 1}}$ & x = ${{3{t^2} + 3} \over {{t^2} - 1}}$

Put    t = tan$\theta$

$\therefore$   x = $-$ 3 sec 2$\theta$  &  2y = 3 ($-$ tan 2$\theta$)

$\because$   sec22$\theta$ $-$ tan22$\theta$ = 1

$\Rightarrow$    ${{{x^2}} \over 9}$ $-$ ${{{y^2}} \over {9/4}}$ = 1

which represents at hyperbola

$\therefore$   a2 = 9  &  b2 = 9/4

$\lambda$(T.A.) = 6; e2 = 1 + ${{9/4} \over 9}$ = 1 + ${1 \over 4}$ $\Rightarrow$ e = ${{\sqrt 5 } \over 2}$
2

### JEE Main 2017 (Online) 8th April Morning Slot

If the common tangents to the parabola, x2 = 4y and the circle, x2 + y2 = 4 intersect at the point P, then the distance of P from the origin, is :
A
$\sqrt 2 + 1$
B
2(3 + 2 $\sqrt 2$)
C
2($\sqrt 2$ + 1)
D
3 + 2$\sqrt 2$

## Explanation Tangent to x2 + y2 = 4 is

y = mx $\pm$ 2$\sqrt {1 + {m^2}}$

Also, x2 = 4y

x2 = 4mx + 8$\sqrt {1 + {m^2}}$

or  x2 = 4mx $-$ 8$\sqrt {1 + {m^2}}$

For D = 0

we have; 16m2 + 4.8$\sqrt {1 + {m^2}}$ = 0

$\Rightarrow$   m2 + 2$\sqrt {1 + {m^2}}$ = 0

$\Rightarrow$    m2 = $-$ 2$\sqrt {1 + {m^2}}$

$\Rightarrow$   m4 = 4 + 4m2

$\Rightarrow$   m4 $-$ 4m2 $-$ 4 = 0

$\Rightarrow$    m2 = ${{4 \pm \sqrt {16 + 16} } \over 2}$

$\Rightarrow$   m2 = ${{4 \pm 4\sqrt 2 } \over 2}$

$\Rightarrow$   m2 = 2 + 2$\sqrt 2$
3

### JEE Main 2017 (Online) 8th April Morning Slot

If two parallel chords of a circle, having diameter 4units, lie on the opposite sides of the center and subtend angles ${\cos ^{ - 1}}\left( {{1 \over 7}} \right)$ and sec$-$1 (7) at the center respectivey, then the distance between these chords, is :
A
${4 \over {\sqrt 7 }}$
B
${8 \over {\sqrt 7 }}$
C
${8 \over 7}$
D
${16 \over 7}$

## Explanation Since cos2$\theta$ = 1/7  $\Rightarrow$ 2 cos2 Q $-$ 1 = 1/7

$\Rightarrow$   2 cos2$\theta$ = 8/7

$\Rightarrow$    cos2 $\theta$ = 4/7

$\Rightarrow$    cos2$\theta$ = ${4 \over 7}$

$\Rightarrow$   cos2$\theta$ = ${2 \over {\sqrt 7 }}$

Also, sec2$\phi$ = 7 = ${1 \over {2{{\cos }^2}\phi - 1}}$ 7

= cos2$\phi$ $-$ 1 = ${1 \over 7}$

= 2 cos2 $\phi$ = ${8 \over 7}$

= cos$\phi$ = ${2 \over {\sqrt 7 }}$

P1P2 = r cos$\theta$ + r cos$\phi$

= ${4 \over {\sqrt 7 }} + {4 \over {\sqrt 7 }}$ = ${8 \over {\sqrt 7 }}$
4

### JEE Main 2017 (Online) 8th April Morning Slot

Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is ${3 \over 5}$ and the distance between its foci is 6, then the area (in sq. units) of the quadrilatateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is :
A
8
B
32
C
80
D
40

## Explanation

e = 3/5 & 2ae = 6  $\Rightarrow$   a = 5

$\because$   b2 = a2 (1 $-$ e2)

$\Rightarrow$   b2 = 25(1 $-$ 9/25) $\Rightarrow$   b = 4

$\therefore$   area of required quadrilateral

= 4(1/2 ab) = 2ab = 40