1

### JEE Main 2021 (Online) 1st September Evening Shift

Let $\theta$ be the acute angle between the tangents to the ellipse ${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$ and the circle ${x^2} + {y^2} = 3$ at their point of intersection in the first quadrant. Then tan$\theta$ is equal to :
A
${5 \over {2\sqrt 3 }}$
B
${2 \over {\sqrt 3 }}$
C
${4 \over {\sqrt 3 }}$
D
2

## Explanation

The point of intersection of the curves ${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$ and ${x^2} + {y^2} = 3$ in the first quadrant is $\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$

Now slope of tangent to the ellipse ${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$ at $\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$ is

${m_1} = - {1 \over {3\sqrt 3 }}$

And slope of tangent to the circle at $\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$ is m2 $= - \sqrt 3$

So, if angle between both curves is $\theta$ then

$\tan \theta = \left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right| = \left| {{{ - {1 \over {3\sqrt 3 }} + \sqrt 3 } \over {1 + \left( { - {1 \over {3\sqrt 3 }}\left( { - \sqrt 3 } \right)} \right)}}} \right|$

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

Option (b)
2

### JEE Main 2021 (Online) 1st September Evening Shift

Consider the parabola with vertex $\left( {{1 \over 2},{3 \over 4}} \right)$ and the directrix $y = {1 \over 2}$. Let P be the point where the parabola meets the line $x = - {1 \over 2}$. If the normal to the parabola at P intersects the parabola again at the point Q, then (PQ)2 is equal to :
A
${{75} \over 8}$
B
${{125} \over {16}}$
C
${{25} \over 2}$
D
${{15} \over 2}$

## Explanation $\left( {y - {3 \over 4}} \right) = {\left( {x - {1 \over 2}} \right)^2}$ .... (1)

For $x = - {1 \over 2}$

$y - {3 \over 4} = 1 \Rightarrow y = {7 \over 4} \Rightarrow P\left( { - {1 \over 2},{7 \over 4}} \right)$

Now, $y' = 2\left( {x - {1 \over 2}} \right)$

At $x = - {1 \over 2}$

$\Rightarrow {m_T} = - 2,{m_N} = {1 \over 2}$

Equation of normal is

$y - {7 \over 4} = {1 \over 2}\left( {x + {1 \over 2}} \right)$

$y = {x \over 2} + 2$

Now put y in equation (1)

${x \over 2} + 2 - {3 \over 4} = {\left( {x - {1 \over 2}} \right)^2}$

$\Rightarrow x = 2$ & $- {1 \over 2}$

$\Rightarrow$ Q(2, 3)

Now, ${(PQ)^2} = {{125} \over {16}}$

Option (b)
3

### JEE Main 2021 (Online) 31st August Evening Shift

An angle of intersection of the curves, ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ and x2 + y2 = ab, a > b, is :
A
${\tan ^{ - 1}}\left( {{{a + b} \over {\sqrt {ab} }}} \right)$
B
${\tan ^{ - 1}}\left( {{{a - b} \over {2\sqrt {ab} }}} \right)$
C
${\tan ^{ - 1}}\left( {{{a - b} \over {\sqrt {ab} }}} \right)$
D
${\tan ^{ - 1}}\left( {2\sqrt {ab} } \right)$

## Explanation

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,{x^2} + {y^2} = ab$

${{2{x_1}} \over {{a^2}}} + {{2{y_1}y'} \over {{b^2}}} = 0$

$\Rightarrow {y_1}' = {{ - {x_1}} \over {{a^2}}}{{{b^2}} \over {{y_1}}}$ .... (1)

$\therefore$ $2{x_1} + 2{y_1}y' = 0$

$\Rightarrow {y_2}' = {{ - {x_1}} \over {{y_1}}}$ ..... (2)

Here (x1y1) is point of intersection of both curves

$\therefore$ $x_1^2 = {{{a^2}b} \over {a + b}},y_1^2 = {{a{b^2}} \over {a + b}}$

$\therefore$ $\tan \theta = \left| {{{{y_1}' - {y_2}'} \over {1 + {y_1}'{y_2}'}}} \right| = \left| {{{{{ - {x_1}{b^2}} \over {{a^2}{y_1}}} + {{{x_1}} \over {{y_1}}}} \over {1 + {{x_1^2{b^2}} \over {{a^2}y_1^2}}}}} \right|$

$\tan \theta = \left| {{{ - {b^2}{x_1}{y_1} + {a^2}{x_1}{y_1}} \over {{a^2}y_1^2 + {b^2}x_1^2}}} \right|$

$\tan \theta = \left| {{{a - b} \over {\sqrt {ab} }}} \right|$
4

### JEE Main 2021 (Online) 31st August Evening Shift

The locus of mid-points of the line segments joining ($-$3, $-$5) and the points on the ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$ is :
A
$9{x^2} + 4{y^2} + 18x + 8y + 145 = 0$
B
$36{x^2} + 16{y^2} + 90x + 56y + 145 = 0$
C
$36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$
D
$36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$

## Explanation

General point on ${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$ is A(2cos$\theta$, 3sin$\theta$)

given B($-$3, $-$5)

midpoint $C\left( {{{2\cos \theta - 3} \over 2},{{3\sin \theta - 5} \over 2}} \right)$

$h = {{2\cos \theta - 3} \over 2};k = {{3\sin \theta - 5} \over 2}$

$\Rightarrow {\left( {{{2h + 3} \over 2}} \right)^2} + {\left( {{{2k + 5} \over 3}} \right)^2} = 1$

$\Rightarrow 36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$