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

The correct option is B

$\frac{9}{2}$

Explanation for the correct option

Step 1: Find the slope

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 $m_1$ & the slope of $9x^2+b{y}^2=16$ be $m_2$.
Thus, we have ${\mathrm{m}}_1{\mathrm{m}}_2=-1$
Differentiating both the equations with respect to $x$ we get:
$$\begin{gathered} \Rightarrow 2y\frac{dy}{dx}=6 \\ \Rightarrow m_1=\frac{3}{y}\dots \dots ..\left(1\right) \end{gathered}$$
And
$$\begin{gathered} \Rightarrow 18x+2by\frac{dy}{dx}=0 \\ \Rightarrow {\mathrm{m}}_2=\frac{-9\mathrm{x}}{\mathrm{by}} \end{gathered}$$

Step 2: Find the value of $b$
We know that ${\mathrm{m}}_1{\mathrm{m}}_2=-1$.
$$\begin{gathered} \Rightarrow \frac{3}{\mathrm{y}}\times \frac{-9\mathrm{x}}{by}=1 \\ \Rightarrow b=\frac{27x}{y^2} \end{gathered}$$
Putting $y^2=6x$ in the above equation we get:
$\begin{array}{rcl}\Rightarrow \mathrm{b}& =& \frac{27\mathrm{x}}{6\mathrm{x}}\\ & =& \frac{{\displaystyle 9}}{{\displaystyle 2}}\end{array}$

Hence, option B is correct.