Question

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 $xy=1-y$, is given by

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

Answer 1

Answer: (C)

Solving curves, $x^{2}y=xy\Rightarrow xy(x-1)=0$
$\Rightarrow x=0,y=0,x=1$
$\because y\ne 0$, so point of intersection of two curves are $(0,1)$ and $(1,1/2)$
$x^{2}y=1-y\Rightarrow x^2\frac{dy}{dx}+2xy=-\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=-\frac{2xy}{x^2+1}$
${\left(\frac{dy}{dx}\right)}_{(0,1)}=0$ 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 $x+2y-2=0$
These intersect at $(0,1)$