A point and an ellipse In particular, I have to find the exact value of the minimum distance from $P(- \frac {15}{4},1)$ to the ellipse $ \frac {x^2}{4} + \frac {y^2}{9} =1$
Therefore, if $ \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ is an ellipse, with the parameterization $x(θ)≔(a \cos ⁡θ,b \sin⁡ θ ),$ I have to find the value of $θ$ giving the minimum distance from $P(p,q)$ (not on the ellipse) to the ellipse is given by a quartic in $t= \tan⁡( \frac {θ}{2}).$
A necessary condition for $x$ to be the closest point to $P$ is that $P-x$ is perpendicular to the tangent vector in $x ,$ i.e. $(P-x(θ) ). x' (θ)=0$
I can't handle the above condition to make an fuction (e.g. $f(θ)$) to find the minimum value by calculating the derivative $f'(θ)=0$ for example. Then I have to prove that rational, non-zero values of $a, b, p, q$ can be found such that the quartic factorises as the product of two quadratics with rational coefficients. Any help? Thank you
 A: Let's solve this step by step:
Parametric equations of ellipse are:
$$x=2\cos\varphi,\quad y=3\sin\varphi$$
The slope of tangent is:
$$y'=\frac{dy}{dx}=\frac{\frac{dy}{d\varphi}}{\frac{dx}{d\varphi}}=-\frac{3\cos\varphi}{2\sin\varphi}$$
Suppose that the point $A$ you are looking has $\varphi=\varphi_1$
The line $AP$ has to be perpendicular to the tangent, which means that the slope of $AP$ has to be equal to $-1/y'_1$. In other words:
$$\frac{1-3\sin\varphi_1}{-\frac{15}{4}-2\cos\varphi_1}=\frac{2\sin\varphi_1}{3\cos\varphi_1}$$
This transforms into equation:
$$6\cos\varphi_1+15\sin\varphi_1-10\sin\varphi_1\cos\varphi_1=0$$.
Introduce subtitution:
$$t=\tan\frac{\varphi_1}{2}, \quad \sin\varphi_1=\frac{2t}{1+t^2}, \quad \cos\varphi_1=\frac{1-t^2}{1+t^2}$$
The equation becomes:
$$3t^4-25t^3-5t-3=0$$
...and this can be factorized as:
$$(t^2-8t-3)(3t^2-t+1)=0$$
So it's either:
$$3t^2-t+1=0$$
...or 
$$t^2-8t-3=0$$
The first equation has no real solutions, you can ignore it completely. And the second one is easy to solve. I think you can take it from here.
