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

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Aicharya · Accepted Answer

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

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

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