Question

The two curves $x=y^2$, $xy=a^3$ cut orthogonally at a point. Then $a^2$ is equal to

• A. $\frac{1}{3}$
• B. $3$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Given curves $x=y^2$ and $xy=a^3$

Point of intersection is $(a^2,a)$

Slope of tangent to curve I at $(a^2,a)=m_1=\frac{a}{2}$

Slope of tangent to curve II at $(a^2,a)=m_2=-\frac{1}{a}$

Since, the two curves cut orthogonally, we will check using options

If $a^2=\frac{1}{3}$, then $m_1=\frac{1}{2\sqrt{3}}$ and $m_2=-\sqrt{3}$

Since $m_1{m}_2\ne -1$. So option A is incorrect.

For option B, $a^2=3$, then $m_1=\frac{\sqrt{3}}{2}$ and $m_2=-\frac{1}{\sqrt{3}}$

Since $m_1{m}_2\ne -1$. So option B is incorrect.

For option C, $a^2=2$, then $m_1=\frac{1}{\sqrt{2}}$ and $m_2=-\frac{1}{\sqrt{2}}$

Since $m_1{m}_2\ne -1$. So option C is incorrect.

For option D, $a^2=\frac{1}{2}$, then $m_1=\frac{1}{2\sqrt{2}}$ and $m_2=-\sqrt{2}$

Since $m_1{m}_2\ne -1$. So option D is correct.