1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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

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

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

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