Finding the coordinates of the farthest point from a point inside an ellipse without knowing a or b?

Question

I was wondering if anyone could help me figure this out. I have an ellipse and the known info is that the major axis is horizontal the center is at (0,0), there is a point P at (0, $\dfrac{3}{2}$), the maximum distance from this point to anywhere on the ellipse is $\sqrt{7}$, and $\dfrac{c}{a}$ = ${\dfrac{\sqrt{3}}{2}}$ (c is the foci length, a is the major axis length). Is there any way to find the equation of this ellipse and the coordinates of the furthest point on the ellipse from P? Any help would be greatly appreciated thank you!

Answer 1

$$\frac{c}{a}=\frac{\sqrt{3}}{2}\to c=\frac{\sqrt{3} a}{2}$$ we know that $$b^2=a^2-c^2\to a^2=4b^2$$ The equation of the ellipse is $$\frac{x^2}{4b^2}+\frac{y^2}{b^2}=1$$ Write the ellipse in parametric form $$E:(x=2b\cos t,y=b\sin t)$$ The distance between the point $P(0,3/2)$ and a generic point of the ellipse is $$EP^2=f(b)=(2b\cos t)^2+\left(b\sin t-\frac32\right)^2=b^2 \sin ^2 t+4 b^2 \cos ^2 t-3 b \sin t+\frac{9}{4}$$ derivative is zero in $[0,2\pi]$ if $$f'(b)= -3 b \cos t (2 b \sin t+1)=0$$ $$ t=\pm\pi/2;\;t=2\pi-\arcsin\left(\frac{1}{2b}\right);\;t=\pi+\arcsin\left(\frac{1}{2b}\right)$$ Looking at the second derivative $f''(b)=3 b (\sin t-2 b \cos 2t)$, if $b>\frac{1}{2}$ the maximum is $t=\pi+\arcsin\left(\frac{1}{2b}\right)$ $$f\left(\pi+\arcsin\left(\frac{1}{2b}\right)\right)=7$$ if $$4 b^2+3=7\to b^2=1$$ The equation of the ellipse is $$\frac{x^2}{4}+y^2=1$$ and the coordinates of the most distant point are $$\left(\pm\sqrt{3},-\frac{1}{2}\right)$$

Edit

You can't have an ellipse with point $P$ internal and $\sqrt 7$ maximum distance. See the second picture below.

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