Question

If the curves $y^2=6x,9x^2+b{y}^2=16$ 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 $\frac{9}{2}$

Let the point of intersection be $(x_1,y_1)$ finding slope of both the curves at point of intersection for $y^2=6x,9x^2+b{y}^2=16$
$2y\frac{dy}{dx}=6,m_1=\frac{6}{2y_1}$
And $18x+2by\frac{dy}{dx}=0,m_2=-\frac{18x_1}{2b{y}_1}$
$m_1{m}_2=-1$
$\Rightarrow \left(\frac{6}{2y_1}\right)\left(\frac{-18x_1}{2b{y}_1}\right)=-1$
$\Rightarrow \frac{\left(6\right)(-18)x_1}{\left(4b\right)\left(6x_1\right)}=-1\Rightarrow b=\frac{9}{2}$