Question

The point of intersection of the tangents drawn to the curve $x^{2}y=1-y$ at the point where it is met by the curve $xy=1-y$ is given by

• A. $\left(2,-1\right)$
• B. $\left(1,2\right)$
• C. $\left(0,1\right)$
• D. None of these

Answer 1

The correct option is C

$\left(0,1\right)$

Find the point of intersection of the tangents drawn to the curve

Given: Equation of curve

$$\begin{gathered} xy=1-y \\ \Rightarrow y=\frac{1}{\left(1+x\right)} \end{gathered}$$

The other curve is $x^{2}y=1-y\to \left(1\right)$

Putting the value of $y$ in equation $\left(1\right)$,

$$\begin{gathered} \Rightarrow x^2\left\{\frac{1}{\left(1+x\right)}\right\}=1-\left\{\frac{1}{\left(1+x\right)}\right\} \\ \Rightarrow \frac{x^2}{1+x}=\frac{x}{1+x} \\ \Rightarrow x=0 \end{gathered}$$

Therefore, $x=0$ and $y=1$.

Hence, option $\left(C\right)$, $\left(0,1\right)$ is the correct answer.