Question

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

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

Hint:

For curves $y^2=ax$ and $Ax^2+By^2=C$ to be orthogonal, the condition is often independent of the specific intersection point.

Answer 1

The Correct Option is D

Step 1: Concept
Curves intersect at right angles (orthogonally) if the product of their slopes at the point of intersection is $-1$ ($m_1 \cdot m_2 = -1$).

Step 2: Meaning
Differentiate both equations to find expressions for their slopes $m_1$ and $m_2$.

Step 3: Analysis
For $y^2 = 6x$: $2y \frac{dy}{dx} = 6 \implies m_1 = \frac{3}{y}$. For $9x^2 + by^2 = 16$: $18x + 2by \frac{dy}{dx} = 0 \implies m_2 = -\frac{9x}{by}$. Orthogonality: $(\frac{3}{y})(-\frac{9x}{by}) = -1 \implies \frac{27x}{by^2} = 1 \implies 27x = by^2$. Substitute $y^2 = 6x$ from the first curve: $27x = b(6x)$.

Step 4: Conclusion
$27 = 6b \implies b = \frac{27}{6} = \frac{9}{2}$. Final Answer: (D)