1

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

Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x is :
A
$2\sqrt 3$y = 12x + 1
B
$\sqrt 3$y = x + 3
C
$2\sqrt 3$y = -x - 12
D
$\sqrt 3$y = 3x + 1

## Explanation

We know,

Equation of tangent to the parabola y2 = 4ax is,

y = mx + ${a \over m}$

$\therefore$  Equation of tangent to the parabola y2 = 4x is,

y = mx + ${1 \over m}$

$\Rightarrow$  m2x $-$ ym + 1 = 0

This tangent is also the tangent to the circle x2 + y2 $-$ 6x = 0

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$\therefore$  $\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$

$\Rightarrow$  (3m2 + 1)2 = 9(m4 + m2)

$\Rightarrow$  9m4 + 6m2 + 1 = 9m4 + 9m2

$\Rightarrow$  3m2 = 1

$\Rightarrow$  m = $\pm$ ${1 \over {\sqrt 3 }}$

So, possible tangents are

y = ${1 \over {\sqrt 3 }}$x + $\sqrt 3$

$\Rightarrow$  $\sqrt 3$y = x + 3

or   y = $-$ ${x \over {\sqrt 3 }}$ $-$ $\sqrt 3$

$\Rightarrow$  $\sqrt 3 y$ = $-$ x $-$ 3
2

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

A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :
A
${3 \over 2}$
B
$\sqrt 3$
C
2
D
${2 \over {\sqrt 3 }}$

## Explanation

Let the equation of hyperbola

${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}}$ = 1

Given 2a = 4

$\Rightarrow$  $a$ = 2

It passes through (4, 2)

$\therefore$  ${{16} \over 4} - {4 \over {{b^2}}}$ = 1

$\Rightarrow$  b2 = ${4 \over 3}$

e = $\sqrt {1 + {{{b^2}} \over {{a^2}}}}$ = $\sqrt {1 + {{4/3} \over 4}}$

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

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

Let A(4, $-$ 4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of $\Delta$ACB is maximum. Then, the area (in sq. units) of $\Delta$ACB, is :
A
$31{1 \over 4}$
B
$30{1 \over 2}$
C
32
D
$31{3 \over 4}$

## Explanation

$\Delta ABC = {1 \over 2}\left| {\matrix{ 4 & { - 4} & 1 \cr 9 & 6 & 1 \cr {{t^2}} & {2t} & 1 \cr } } \right|$

D = 60 + 10t $-$ 10t2

${{d\Delta } \over {dt}} = 0 \Rightarrow t = {1 \over 2}$

${{{d^2}\Delta } \over {d{t^2}}} = - 20 < 0$

$\therefore$  max at $t = {1 \over 2}$

max area $\Delta = 65 - {5 \over 2}$

$= {{125} \over 2} = 31{1 \over 4}$
4

### 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).