1

### JEE Main 2019 (Online) 10th January Evening Slot

The length of the chord of the parabola x2 $=$ 4y having equation x – $\sqrt 2 y + 4\sqrt 2 = 0$  is -
A
$8\sqrt 2$
B
$6\sqrt 3$
C
$3\sqrt 2$
D
$2\sqrt {11}$

## Explanation

x2 = 4y

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

Solving together we get

x2 = 4$\left( {{{x + 4\sqrt 2 } \over {\sqrt 2 }}} \right)$

$\sqrt 2$x2 + 4x + 16$\sqrt 2$

$\sqrt 2$x2 $-$ 4x $-$ 16$\sqrt 2$ = 0

x1 + x2 = 2$\sqrt 2$; x1x2 = ${{ - 16\sqrt 2 } \over {\sqrt 2 }}$ = $-$ 16

Similarly,

($\sqrt 2$y $-$ 4$\sqrt 2$)2 = 4y

2y2 + 32 $-$ 16y = 4y  $\ell$AB = $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

$= \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)}$

$= \sqrt {8 + 64 + 100 - 64}$

$= \sqrt {108} = 6\sqrt 3$
2

### JEE Main 2019 (Online) 10th January Evening Slot

Let S = $\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.$ Then S represents
A
an ellipse whose eccentricity is ${1 \over {\sqrt {r + 1} }},$ where r > 1
B
an ellipse whose eccentricity is ${2 \over {\sqrt {r + 1} }},$ where 0 < r < 1
C
an ellipse whose eccentricity is ${2 \over {\sqrt {r - 1} }},$ where 0 < r < 1
D
an ellipse whose eccentricity is ${2 \over {\sqrt {r + 1} }},$ where r > 1

## Explanation

${{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}} = 1$

for r > 1,     ${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$

$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)}$

$= \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}}$

$= \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}}$
3

### JEE Main 2019 (Online) 11th January Morning Slot

Equation of a common tangent to the parabola y2 = 4x and the hyperbola xy = 2 is
A
x + y + 1 = 0
B
4x + 2y + 1 = 0
C
x – 2y + 4 = 0
D
x + 2y + 4 = 0

## Explanation

Let the equation of tangent to parabola

y2 = 4x be y = mx + ${1 \over m}$

It is also a tangent to hyperbola xy = 2

$\Rightarrow$  x$\left( {mx + {1 \over m}} \right)$ = 2

$\Rightarrow$  x2m + ${x \over m}$ $-$ 2 = 0

D = 0 $\Rightarrow$   m = $-$ ${1 \over 2}$

So tangent is 2y + x + 4 = 0
4

### JEE Main 2019 (Online) 11th January Morning Slot

If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :
A
${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$
B
${1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$
C
${1 \over {4{x^2}}} + {1 \over {2{y^2}}} = 1$
D
${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$

## Explanation

Equation of general tangent on ellipse

${x \over {a\,\sec \theta }} + {y \over {b\cos ec\theta }} = 1$

$a = \sqrt 2 ,\,\,b = 1$

$\Rightarrow {x \over {\sqrt 2 \sec \theta }} + {y \over {\cos ec\theta }} = 1$

Let the midpoint be (h, k)

$h = {{\sqrt 2 \sec \theta } \over 2} \Rightarrow \cos \theta = {1 \over {\sqrt 2 h}}$

and $k = {{\cos ec\theta } \over 2} \Rightarrow \sin \theta = {1 \over {2k}}$

$\because$  ${\sin ^2}\theta + {\cos ^2}\theta = 1$

$\Rightarrow$  ${1 \over {2{h^2}}} + {1 \over {4{k^2}}} = 1$

$\Rightarrow$  ${1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$