Question

If the curves $2x=y^2$ and $2xy=K$ intersect perpendicularly, then the value of $K^2$ is

• A. $4$
• B. $2\sqrt{2}$
• C. $2$
• D. $8$

Answer 1

Answer: (D)

$2x=y^2\dots (1)$
$2xy=K\dots (2)$
Solving $(1)$ and $(2)$, we get
$(x,y)=\left(K^{(2/3)}/2,K^{(1/3)}\right)$
Differentiating $(1)$ and $(2)$ w.r.t. $x$
$m_1=dy/dx=1/y\dots (3)$
$m_2=dy/dx=-y/x\dots (4)$
Both curves intersect each other perpendicularly
$\therefore m_1{m}_2=-1$
$\Rightarrow -1/x=-1$
$\Rightarrow x=1$
$\Rightarrow K^{(2/3)}=2$
$\Rightarrow K^2=8$