QJEEM20Ma8115: JEE Main 2020 Mathematics - Class 12 JEEM20Ma8115 Solution: Let y = y(x) be a function of x satisfying where k is a…

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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}

Step-by-Step Verified Solution