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

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Aicharya · Accepted Answer

$\begin{aligned} & y^2=6 x ..(i)\ & \Rightarrow 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=6 \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{3}{y} \ & 	ext { Also, } 9 x^2+\mathrm{b} y^2=16 \ & \Rightarrow 18 x+2 \mathrm{~b} y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0 \ & \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-9 x}{\mathrm{~b} y} \end{aligned}$ As given curves intersect each other at right angle, their tangents also intersect at right angles. $\begin{array}{ll} & \frac{3}{y} 	imes \frac{-9 x}{b y}=-1 \ \Rightarrow \quad & \Rightarrow y^2=27 x \ & 	ext { (i) } \Rightarrow b(6 x)=27 x \ & \Rightarrow b=\frac{9}{2}\end{array}$

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

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