JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 14)
Let y = y(x) be a function of x satisfying
$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ where k is a constant and
$$y\left( {{1 \over 2}} \right) = - {1 \over 4}$$. Then $${{dy} \over {dx}}$$ at x = $${1 \over 2}$$, is equal to :
$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ where k is a constant and
$$y\left( {{1 \over 2}} \right) = - {1 \over 4}$$. Then $${{dy} \over {dx}}$$ at x = $${1 \over 2}$$, is equal to :
$${2 \over {\sqrt 5 }}$$
$$ - {{\sqrt 5 } \over 2}$$
$${{\sqrt 5 } \over 2}$$
$$ - {{\sqrt 5 } \over 4}$$
Explanation
$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ ....(1)
On differentiating both side of eq. (1) w.r.t. x we get,
$${{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}$$
= 0 - $$\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}$$
Put x = $${1 \over 2}$$ and y = $$ - {1 \over 4}$$, we get
$${{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}$$
= $$ - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}$$
$$ \therefore $$ $${{dy} \over {dx}} = - {{\sqrt 5 } \over 2}$$
On differentiating both side of eq. (1) w.r.t. x we get,
$${{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}$$
= 0 - $$\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}$$
Put x = $${1 \over 2}$$ and y = $$ - {1 \over 4}$$, we get
$${{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}$$
= $$ - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}$$
$$ \therefore $$ $${{dy} \over {dx}} = - {{\sqrt 5 } \over 2}$$
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