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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

Steven
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  • This is similar to your previous question Here it is – tien lee Jun 23 '18 at 21:16
  • you are right, but I haven't got an answer as well – Steven Jun 23 '18 at 21:17
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    You got an answer with an explanation as well – tien lee Jun 23 '18 at 21:23
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    It was not simple for me, I am sure it was a good answer, but my goal is to understand the solution. Thank you for your notices – Steven Jun 23 '18 at 21:25
  • It’s not just similar to your previous question. It’s virtually identical to it. If the answers that you’ve already gotten aren’t satisfactory, wait for others or add more context to that question so that potential answerers have a better idea of what you’re looking for. Don’t post a new question identical to the original one with a couple of specific numbers plugged in. – amd Jun 24 '18 at 04:36

1 Answers1

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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.

Oldboy
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