A) \[1/2\]
B) \[1\]
C) \[2\]
D) None of these
Correct Answer: B
Solution :
Given curves, \[\frac{{{x}^{2}}}{{{a}^{3}}}+\frac{{{y}^{2}}}{2}=1\] ?..(i) On differentiating, \[\frac{2x}{{{a}^{2}}}+\frac{2y}{2}\,\,\frac{dy}{dx}=0\] \[\Rightarrow \] \[y\frac{dy}{dx}=\frac{-2x}{{{d}^{2}}}\] Let \[{{m}_{1}}=\frac{dy}{dx}=\frac{-2x}{{{d}^{2}}y}\] ?..(ii) and the curve, \[{{y}^{2}}=8x\] On differentiating, \[2y\frac{dy}{dx}=8\] \[\Rightarrow \] Let \[{{m}_{2}}=\frac{dy}{dx}=\frac{4}{y}\] ?.(iii) \[\because \] Both curves intersect at right angles \[\Rightarrow \] \[{{m}_{1}}{{m}_{2}}=-1\] \[\Rightarrow \] \[-\frac{2x}{{{a}^{2}}y}\times \frac{4}{y}=-1\] \[\Rightarrow \] \[{{a}^{2}}=\frac{8x}{{{y}^{2}}}\] \[\Rightarrow \] \[{{a}^{2}}=\frac{8x}{8x}=1\] \[(\because \,\,{{y}^{2}}=8x)\]
You need to login to perform this action.
You will be redirected in
3 sec
