Question

If the curve \(y^2 = 6x\) and \(9x^2 + by^2 = 16\) intersect each other at right angles, then the value of \(b\) is:

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

Hint:

The orthogonality condition \(m_1 \cdot m_2 = -1\) is critical for perpendicular intersections.

Answer 1

The Correct Option is A

We are given two curves: \[ y^2 = 6x \quad \text{(1)} \] and \[ 9x^2 + by^2 = 16 \quad \text{(2)}. \] To find the value of \(b\), we first differentiate both equations with respect to \(x\). From equation (1): \[ 2y \frac{dy}{dx} = 6 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{3}{y}. \] From equation (2): \[ 18x + 2by \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{9x}{by}. \] For the curves to intersect at right angles, the product of the slopes of the tangents at the point of intersection must be \(-1\): \[ \frac{3}{y} \cdot \left(-\frac{9x}{by}\right) = -1. \] Simplifying: \[ -\frac{27x}{by^2} = -1 \quad \Rightarrow \quad \frac{27x}{by^2} = 1. \] Using equation (1), \(y^2 = 6x\), substitute this into the above equation: \[ \frac{27x}{b \cdot 6x} = 1 \quad \Rightarrow \quad \frac{27}{6b} = 1 \quad \Rightarrow \quad b = \frac{9}{2}. \] Thus, the value of \(b\) is \(\frac{9}{2}\).