Ellipse

Let $a^2 = 14$ and $b^2 = 5$ , then,

$P = ( a \cos A, b \sin A )$ , and $Q = ( a \cos 2 A , b \sin 2 A )$

The tangent to the ellipse at P has a direction vector $V = ( - a \sin A, b \cos A )$

Therefore, we need to solve the following equation:

$(Q - P) \cdot V = 0$

Substituting for the variables from above, we obtain,

$a ( \cos 2 A - \cos A ) ( - a \sin A ) + b ( \sin 2 A - \sin A ) ( b \cos A ) = 0$

Since $A \ne 0$ , we can divide by $\sin A$ ,

$- a^2 ( \cos 2 A - \cos A ) + b^2 ( 2 \cos A - 1 ) \cos A = 0$

This becomes,

$- 14 ( 2 \cos^2 A - \cos A - 1 ) + 5 ( 2 \cos^2 A - \cos A ) = 0$

which simplifies to,

$-18 \cos^2 A + 9 \cos A + 14 = 0$

Negating,

$18 \cos^2 A - 9 \cos A - 14 = 0$

which is factorable as

$(6 \cos A -7)(3 \cos A +2 )= 0$

Hence, $\cos A = - \dfrac{2}{3}$