Question

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

• A. \(\frac92\)
• B. 6
• C. \(\frac72\)
• D. 4

Hint:

At a common point the product of the slopes of the two curves must be \(-1\); use \(y^2=6x\) to eliminate \(y\).

Answer 1

The Correct Option is A

Let \((x_1,y_1)\) be a common point. From \(y^2=6x\): \(2y\,y'=6\Rightarrow m_1=\frac{3}{y_1}\).
From \(9x^2+by^2=16\): \(18x+2by\,y'=0\Rightarrow m_2=-\frac{9x_1}{by_1}\).
Orthogonal: \(m_1m_2=-1\Rightarrow-\frac{27x_1}{by_1^2}=-1\Rightarrow by_1^2=27x_1\).
With \(y_1^2=6x_1\): \(6bx_1=27x_1\Rightarrow b=\frac{27}{6}=\frac92\).
(For \(b=\frac92\) the curves do meet at a point with \(x_1\gt 0\): \(9x^2+27x-16=0\) has a positive root.)
Answer: (1) \(\frac92\)