If the line xcos alpha + ysin alpha = p be no | vedclass

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(d) The equation of any normal to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is $ax\,\,\sec \phi - by\,\,{\rm{cosec}}\,\phi = {a^2} - {b^

Vedclass · Accepted Answer

${p^2}({a^2}{\sec ^2}\alpha + {b^2}{\rm{cose}}{{\rm{c}}^2}\alpha ) = {({a^2} - {b^2})^2}$

Vedclass · Answer

(d) The equation of any normal to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is $ax\,\,\sec \phi - by\,\,{\rm{cosec}}\,\phi = {a^2} - {b^2}$.....$(i)$

The straight line $x\cos \alpha + y\sin \alpha = p$ will be a normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

If $(i)$ and $x\cos \alpha + y\sin \alpha = p$ represent the same line $\frac{{a\sec \phi }}{{\cos \alpha }} = \frac{{ - b\,{\rm{cosec}}\phi }}{{\sin \alpha }} = \frac{{{a^2} - {b^2}}}{p}$

$ \Rightarrow \,\cos \phi \, = \frac{{ap}}{{({a^2} - {b^2})\cos \alpha }},\,$$\sin \phi = \frac{{ - bp}}{{({a^2} - {b^2})\sin \alpha }}$

${\sin ^2}\phi + {\cos ^2}\phi = 1$

$ \Rightarrow \,\frac{{{b^2}{p^2}}}{{{{({a^2} - {b^2})}^2}{{\sin }^2}\alpha }} + \frac{{{a^2}{p^2}}}{{{{({a^2} - {b^2})}^2}{{\cos }^2}\alpha }} = 1$

$ \Rightarrow $${p^2}({b^2}{\rm{cose}}{{\rm{c}}^{\rm{2}}}\,\alpha + {a^2}{\sec ^2}\alpha ) = {({a^2} - {b^2})^2}$.

If the line $x\cos \alpha + y\sin \alpha = p$ be normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then

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