Question

If the curves $y^2-2x,9x^2+b{y}^2-16$ intersect each other at right angles, then the value of $b$ is

• A. $\frac{9}{2}$
• B. $6$
• C. $\frac{7}{2}$
• D. $4$

Answer 1

The correct option is A $\frac{9}{2}$

$y^2-2x=0.....\left(1\right)$

$9x^2+b{y}^2-16=0.....\left(2\right)$

Slope of curve $\left(1\right)$-

$2y\frac{dy}{dx}-2=0$

$\frac{dy}{dx}=\frac{1}{y}$

Slope of curve $\left(2\right)$-

$18x+2by\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-9x}{by}$

As both the curves intersect at right angle.

Therefore,

$\left(\frac{1}{y}\right)\left(\frac{-9x}{by}\right)=-1$

$\frac{9x}{b{y}^2}=1$

$9x=b{y}^2$

$9x=b\left(2x\right)\left(\text{From}\left(1\right)\right)$

$\Rightarrow b=\frac{9}{2}$