A) \[\frac{1}{9}\]
B) \[\frac{1}{3}\]
C) 3
D) 18
E) 9
Correct Answer: E
Solution :
Equation of curves are \[4{{x}^{2}}+p{{y}^{2}}=45\]and\[{{x}^{2}}-4{{y}^{2}}=5\] Let\[(\alpha ,\beta )\]be the point of contact. So, from equations of curves, we get \[\frac{{{\alpha }^{2}}}{{{\beta }^{2}}}=\frac{p+36}{5}\] Now, on differentiation of equation of curves w.r.t.\[x,\]we get \[{{m}_{1}}={{\left( \frac{dy}{dx} \right)}_{(\alpha ,\beta )}}=-\frac{4\alpha }{p\beta }\] And \[{{m}_{2}}={{\left( \frac{dy}{dx} \right)}_{(\alpha ,\beta )}}=\frac{\alpha }{4\beta }\] \[\because \]Both curves cut orthogonally, then \[\left( -\frac{4\alpha }{p\beta } \right).\frac{\alpha }{4\beta }=-1\] \[\Rightarrow \] \[\frac{1}{p}\left( \frac{p+36}{5} \right)=1\] \[\Rightarrow \] \[5p=p+36\] \[\Rightarrow \] \[p=9\]
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