If the normal to the ellipse 3x ^ 2 + 4 y ^ 2 = 12 at a point P on...

Given the equation of an ellipse is $\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{3} = 1$,where $a = 2, b = \sqrt{3} .$
As we know equation of normal to the ellipse $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ can be taken as
$axsec \theta - bycosec \theta = {a}^{2} - {b}^{2}$
$⇒ 2xsec \theta - \sqrt{3}ycosec \theta = 1$ for the given ellipse.
Now, this normal is parallel to $2x + y = 5$.
$⇒ - \dfrac{2sec \theta}{\sqrt{3}cosec \theta} = - 2 ⇒ tan \theta = - \sqrt{3} ⇒ \theta = \dfrac{2\pi}{3}, - \dfrac{\pi}{3}$
Hence point $P\left(acos \theta, bsin \theta\right) = \left(- 1, \dfrac{3}{2}\right) or \left(1, - \dfrac{3}{2}\right)$
Tangent at \left(- 1, \dfrac{3}{2}\right)$is$\dfrac{x . \left(- 1\right)}{4} + \dfrac{y . \dfrac{3}{2}}{3} = 1$<br />$⇒ - x + 2y = 4$which also passes through$Q\left(4, 4\right)$<br />$⇒ PQ = \sqrt{{\left(4 \pm 1\right)}^{2} + {\left(4 ∓ \dfrac{3}{2}\right)}^{2}} = \dfrac{5\sqrt{5}}{2} .