Q95 / 200 · AP EAMCET 2002 · Mathematics
The two curves $x=y^2, \quad x y=a^3 \quad$ cut orthogonally at a point, then $a^2$ is equal to
- A$\frac{1}{3}$
- Correct answerB$\frac{1}{2}$
- C$1$
- D$2$
Solution
We have,
$$
x=y^2
$$
$$
\Rightarrow \quad 2 y \frac{d y}{d x}=1 \Rightarrow \frac{d y}{d x}=\frac{1}{2 y}
$$
and
$$
x y=a^3
$$
$$
\Rightarrow \quad x \frac{d y}{d x}+y=0 \Rightarrow \frac{d y}{d x}=-\frac{y}{x}
$$
On solving equations (i) and (ii), we get the point of intersection $\left(a^2, a\right)$
$$
\begin{aligned}
m_1 & =\left(\frac{d y}{d x}\right)_{\left(a^2, a\right)}=\frac{1}{2 a^{\prime}} \\
\text { and } \quad m_2 & =\left(\frac{d y}{d x}\right)_{\left(a^2, a\right)}=\frac{-a}{a^2}=-\frac{1}{a}
\end{aligned}
$$
Since, curves cut orthogonally then,
$$
\begin{array}{lrl}
m_1 m_2=-1 & \\
\Rightarrow & \left(\frac{1}{2 a}\right)\left(-\frac{1}{a}\right) & =-1 \\
\Rightarrow & a^2 & =\frac{1}{2}
\end{array}
$$
Hence, the correct answer is option: 'B'.