Question

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 $xy=1-y$, is given by

• A. $(0,-1)$
• B. $(1,1)$
• C. $(0,1)$
• D. None of these

Answer 1

Answer: (C)

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, $x^{2}y=1-y$
$\Rightarrow x^2\left(\frac{dy}{dx}\right)+2xy=-\left(\frac{dy}{dx}\right)$
$\Rightarrow \frac{dy}{dx}=-\frac{2xy}{x^2+1}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(0,1)}=0$ 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 $x+2y-2=0$
These two tangents intersect at $(0,1)$.