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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 met by the curve $x y=1-y$, is given by
Application of Derivatives
Solution:
Solving the two equations, we get
$x ^{2} y = xy $
$\Rightarrow xy ( x -1)=0 $
$\Rightarrow x =0, y =0, x =1$
Since $y=0$ does not satisfy the two equations,
So, we neglect it.
Putting $x =0$ in the either equation, we get $y =1$.
Now, putting $x =1$ in one of the two equations
we obtain $y =\frac{1}{2}$.
Thus, the two curve interset at $(0,1)$ and $\left(1, \frac{1}{2}\right)$.
Now, $x^{2} y=1-y $
$\Rightarrow x^{2}\left(\frac{d y}{d x}\right)+2 x y=-\left(\frac{d y}{d x}\right)$
$\Rightarrow \frac{ dy }{ dx }=-\frac{2 xy }{ x ^{2}+1}$
$ \Rightarrow \left(\frac{ dy }{ dx }\right)_{(0,1)}=0$ and
$\left(\frac{ dy }{ dx }\right)_{(1,1 / 2)}=-\frac{1}{2}$.
The equations of the required tangents are
$y-1=0(x-0)$ and $y-1 / 2=1 / 2(x-1)$
$\Rightarrow y =1$ and $x +2 y -2=0$
These two tangents intersect at $(0,1)$.