Question

If the curve $y^2=6x,9x^2+b{y}^2=16$ intersect each other at right angles, then the value of $b$ is

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

Answer 1

Answer: (B)

$\frac{9}{2}$

It is given that the curves $y^2=6x$ and $9x^2+b{y}^2=16$ intersect at right angles.

Let the slope of $y^2=6x$ be $m_1$

Also, let the slope of $9x^2+b{y}^2=16$ be $m_2$

Thus, we have $m_1{m}_2=-1$

Differentiating both the equations wrt $x$ we get

$2y\frac{dy}{dx}=6$

Or, $y{m}_1=3$........$\left(1\right)$

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

Or, $m_2=\frac{-9x}{by}$

We know that $m_1{m}_2=-1$

Thus, $\frac{3}{y}\times \frac{-9x}{by}=1$

$\to b=\frac{27x}{y^2}$

Putting $y^2=6x$ in the above equation we get

$b=\frac{27x}{6x}=\frac{9}{2}$