1

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

If the parabolas y2 = 4b(x – c) and y2 = 8ax have a common normal, then which on of the following is a valid choice for the ordered triad (a, b, c) ?
A
(1, 1, 3)
B
(1, 1, 0)
C
$\left( {{1 \over 2},2,0} \right)$
D
$\left( {{1 \over 2},2,3} \right)$

## Explanation

Normal to the two given curves are

y = m(x – c) – 2bm – bm3,

y = mx – 4am – 2am3

If they have a common normal, then

(c + 2b)m + bm3 = 4am + 2am3

$\Rightarrow$ (4a – c – 2b) m = (b – 2a)m3

$\Rightarrow$ (4a – c – 2b) = (b – 2a)m2

$\Rightarrow$ m2 = ${c \over {2a - b}} - 2$ $>$ 0

By checking all options we found (A) is right option.

Note : We get that all the options are correct for m = 0 i.e., when common normal is x-axis. But may be in question they want common normal other than x – axis, hence answer is (A).
2

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

The equation of a tangent to the hyperbola 4x2 – 5y2 = 20 parallel to the line x – y = 2 is -
A
x $-$ y + 9 = 0
B
x $-$ y $-$ 3 = 0
C
x $-$ y + 1 = 0
D
x $-$ y + 7 = 0

## Explanation

Hyperbola ${{{x^2}} \over 5} - {{{y^2}} \over 4} = 1$

slope of tangent = 1

equation of tangent y = x $\pm$ $\sqrt {5 - 4}$

$\Rightarrow$  y = x $\pm$ 1

$\Rightarrow$  y = x + 1

or

$\Rightarrow$  y = x $-$ 1
3

### 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 = $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

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

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

$= \sqrt {108} = 6\sqrt 3$
4

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

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

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

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