If the normal at the point P(theta ) to the e | vedclass

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Question

(b) The normal at $P(a\cos 	heta ,b\sin 	heta )$ is $ax\sec 	heta - by\,\,{m{cosec}}	heta = {a^2} - {b^2}$, where ${a^2} = 14,\,\,{b^2} = 5$

Vedclass · Accepted Answer

$ - \frac{2}{3}$

Vedclass · Answer

(b) The normal at $P(a\cos 	heta ,b\sin 	heta )$ is $ax\sec 	heta - by\,\,{m{cosec}}	heta = {a^2} - {b^2}$, where ${a^2} = 14,\,\,{b^2} = 5$

It meets the curve again at $Q(2	heta )$

$i.e.$ ,$(a\cos 2	heta ,\,b\sin 2	heta )$.

$	herefore \frac{a}{{\cos 	heta }}a\cos 2	heta - \frac{b}{{\sin 	heta }}(b\sin 2	heta ) = {a^2} - {b^2}$

$ \Rightarrow \frac{{14}}{{\cos 	heta }}\cos 2	heta - \frac{5}{{\sin 	heta }}(\sin 2	heta ) = 14 - 5$

$ \Rightarrow 18{\cos ^2}	heta - 9\cos 	heta - 14 = 0$

$ \Rightarrow (6\cos 	heta - 7)(3\cos 	heta + 2) = 0 $

$\Rightarrow \cos 	heta = - \frac{2}{3}$.

If the normal at the point $P(\theta )$ to the ellipse $\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1$ intersects it again at the point $Q(2\theta )$, then $\cos \theta $ is equal to

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