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Q.
The point of intersection of the tangents drawn to the curve $x^{2} y=1-y$ at the points where it is meets the curve $x y=1-y$, is given by
Application of Derivatives
Solution:
Solving curves, $x^{2} y=xy \Rightarrow x y(x-1)=0$
$ \Rightarrow x=0, y=0, x=1$
$\because y \neq 0$, so point of intersection of two curves are $(0,1)$ and $(1,1 / 2)$
$x^{2} y=1-y \Rightarrow x^{2} \frac{d y}{d x}+2 x y=-\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=-\frac{2xy}{x^{2}+1}$
$\left(\frac{dy}{dx}\right)_{(0,1)}=0$ and $\left(\frac{dy}{dx}\right)_{(1,1 / 2)}=-\frac{1}{2}$
Equations of tangents are
$(y-1)=0(x-0)$ and $y-1 / 2=-1 / 2(x-1)$
$y=1$ and $x+2 y-2=0$
These intersect at $(0,1)$