If the normal at $(4 \cos \alpha, 3 \sin \alpha)$ on the ellipse $9 x^2+16 y^2=144$ again intersects the ellipse at $(4 \cos \beta, 3 \sin \beta)$, then $\cos \beta=\cos \alpha\left(\frac{a-b \cos ^2 \alpha}{c-d \cos ^2 \alpha}\right)$ where $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathrm{N}$ and $\operatorname{GCD}(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d})=1 .(\alpha \neq \mathrm{n} \pi, \mathrm{n} \in \mathrm{I})$ we have to find ab,c,d.
I have written equation of normal to ellipse at first point. Then substituted second point and got a relationbut stuck after that
Let $P = (x_1, y_1) = ( 4 \cos \alpha, 3 \sin \alpha ) $
The normal vector at $P$ is $ g = ( 3 \cos \alpha, 4 \sin \alpha)$
Therefore, the normal line at $P$ has the parametric equation
$ \ell(s) = P + s g = (4 \cos \alpha , 3 \sin \alpha) + s ( 3 \cos \alpha, 4 \sin \alpha) = ( (4 + 3 s) \cos \alpha , (3 + 4 s ) \sin \alpha )$
Substituting this into the equation of the ellipse, we get
$ 9 (4 + 3 s)^2 \cos^2 \alpha + 16 (3 + 4 s)^2 \sin^2 \alpha = 144 $
And this reduces to
$ s( 216 \cos^2 \alpha + 224 \sin^2 \alpha ) + s^2 ( 81 \cos^2 \alpha + 256 \sin^2 \alpha) = 0 $
Since $s\ne 0$ then
$s = - \dfrac{ 216 \cos^2 \alpha + 224 \sin^2 \alpha }{ 81 \cos^2 \alpha + 256 \sin^2 \alpha } $
Substitute this into the expression for $\ell(s)$, you get
$ Q = P + s g = ( (4 + 3 s ) \cos \alpha, (3 + 4 s ) \sin \alpha ) \\ = \dfrac{( (-324 \cos^2 \alpha + 352 \sin^2 \alpha) \cos \alpha , (-621 \cos^2 \alpha - 128 \sin^2 \alpha ) \sin \alpha ) }{81 \cos^2 \alpha + 256 \sin^2 \alpha }$
But $Q = ( 4 \cos \beta , 3 \sin \beta ) $
Therefore,
$ \cos \beta = \cos \alpha \left( \dfrac{ -81 \cos^2 \alpha + 88 \sin^2 \alpha }{81 \cos^2 \alpha + 256 \sin^2 \alpha }\right) $
Changing $\sin^2 \alpha$ into $1 - \cos^2 \alpha$, we get:
$ \cos \beta = \cos \alpha \left( \dfrac{ 88 - 169 \cos^2 \alpha}{256 - 175 \cos^2 \alpha} \right) $
Therefore, $a = 88 , b = 169, c = 256, d = 175 $