Question

If the curves $ \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{12}=1 $ and $ {{y}^{3}}=8x $ intersect at right angle, then the value of $ {{a}^{2}} $ is equal to

• A. 16
• B. 12
• C. 8
• D. 4
• E. 2

Answer 1

Answer: (D)

Given curves are $ \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{12}=1 $
$ \Rightarrow $ $ \frac{2x}{{{a}^{2}}}+\frac{2y}{12}.\frac{dy}{dx}=0 $
$ \Rightarrow $ $ \frac{dy}{dx}=-\frac{12x}{{{a}^{2}}y}={{m}_{1}} $
(say) and $ {{y}^{3}}=8x\Rightarrow 3{{y}^{2}}\frac{dy}{dx}=8 $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{8}{3{{y}^{2}}}={{m}_{2}} $ (say) For $ \theta =\frac{\pi }{2},1+{{m}_{1}}{{m}_{2}}=0 $
$ \Rightarrow $ $ 1+\left( \frac{-12x}{{{a}^{2}}y} \right)\left( \frac{8}{3{{y}^{2}}} \right)=0 $
$ \Rightarrow $ $ 3{{a}^{2}}(8x)-96x=0 $
$ \Rightarrow $ $ {{a}^{2}}=4 $