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(B) The equation of the normal at point $P(a\cos 	heta, b\sin 	heta)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $ax \sec

Vedclass · Accepted Answer

$-\frac{2}{3}$

Vedclass · Answer

(B) The equation of the normal at point $P(a\cos 	heta, b\sin 	heta)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $ax \sec 	heta - by \csc 	heta = a^2 - b^2$.Given $a^2 = 14$ and $b^2 = 5$,the normal equation is $\frac{14x}{a\cos 	heta} - \frac{5y}{b\sin 	heta} = 14 - 5 = 9$.Substituting the coordinates of point $Q(a\cos 2	heta, b\sin 2	heta)$ into the normal equation:$\frac{14(a\cos 2	heta)}{a\cos 	heta} - \frac{5(b\sin 2	heta)}{b\sin 	heta} = 9$$14 \frac{\cos 2	heta}{\cos 	heta} - 5 \frac{\sin 2	heta}{\sin 	heta} = 9$Using $\cos 2	heta = 2\cos^2 	heta - 1$ and $\sin 2	heta = 2\sin 	heta \cos 	heta$:$14 \frac{2\cos^2 	heta - 1}{\cos 	heta} - 5 \frac{2\sin 	heta \cos 	heta}{\sin 	heta} = 9$$28\cos 	heta - \frac{14}{\cos 	heta} - 10\cos 	heta = 9$$18\cos 	heta - \frac{14}{\cos 	heta} = 9$$18\cos^2 	heta - 9\cos 	heta - 14 = 0$Factoring the quadratic equation:$(6\cos 	heta - 7)(3\cos 	heta + 2) = 0$Since $|\cos 	heta| \leq 1$,we have $3\cos 	heta + 2 = 0$,which gives $\cos 	heta = -\frac{2}{3}$.

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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