JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 15)
If the curve x2 + 2y2 = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is :
$${\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 4}} \right)$$
$${\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$$
$${\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$$
$${\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)$$
Explanation
Ellipse : $${x \over 2} + {y \over 1} = 1$$
Line : $$x + y = 1$$
_25th_February_Evening_Shift_en_15_1.png)
Using homogenisation
$${x^2} + 2{y^2} = 2{(1)^2}$$
$${x^2} + 2{y^2} = 2{(x + y)^2}$$
$${x^2} + 2{y^2} = 2{x^2} + 2{y^2} + 4xy$$
$${x^2} + 4xy = 0$$
for $$a{x^2} + 2hxy + b{y^2} = 0$$
$$\tan \theta = \left| {{{2\sqrt {{h^2} - ab} } \over {a + b}}} \right|$$
$$\tan \theta = \left| {{{2\sqrt {{{(2)}^2} - 0} } \over {1 + 0}}} \right|$$
$$\tan \theta = - 4$$
$$\cot \theta = - {1 \over 4}$$
$$\theta = {\cot ^{ - 1}}\left( { - {1 \over 4}} \right)$$
$$\theta = \pi - {\cot ^{ - 1}}\left( {{1 \over 4}} \right)$$
$$\theta = \pi - \left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {{1 \over 4}} \right)} \right)$$
$$\theta = {\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)$$
Line : $$x + y = 1$$
Using homogenisation
$${x^2} + 2{y^2} = 2{(1)^2}$$
$${x^2} + 2{y^2} = 2{(x + y)^2}$$
$${x^2} + 2{y^2} = 2{x^2} + 2{y^2} + 4xy$$
$${x^2} + 4xy = 0$$
for $$a{x^2} + 2hxy + b{y^2} = 0$$
$$\tan \theta = \left| {{{2\sqrt {{h^2} - ab} } \over {a + b}}} \right|$$
$$\tan \theta = \left| {{{2\sqrt {{{(2)}^2} - 0} } \over {1 + 0}}} \right|$$
$$\tan \theta = - 4$$
$$\cot \theta = - {1 \over 4}$$
$$\theta = {\cot ^{ - 1}}\left( { - {1 \over 4}} \right)$$
$$\theta = \pi - {\cot ^{ - 1}}\left( {{1 \over 4}} \right)$$
$$\theta = \pi - \left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {{1 \over 4}} \right)} \right)$$
$$\theta = {\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)$$
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