Question

Find the lengths of, and the equations to, the focal radii drawn to the point $(4\sqrt{3},5)$ of the ellipse $25x^2+16y^2=1600$.

Answer 1

$25x^2+16y^2=1600\frac{25x^2}{1600}+\frac{16y^2}{1600}=1\frac{x^2}{64}+\frac{{\left(y\right)}^2}{100}=1a=8,b=10b>a$

So the major axis of ellipise is $y$ axis.

$e^2=1-\frac{a^2}{b^2}=1-\frac{64}{100}=\frac{36}{100}\Rightarrow e=\frac{6}{10}$

Foci of the ellipse are $(0,\pm be)$

$(0,\pm 10\times \frac{6}{10})\Rightarrow (0,\pm 6)$

$S(0,6)$ and $S^{\prime }(0,-6)$

Given point on ellipse is $P(4\sqrt{3},5)$

Equation of $PS$

$y-6=\frac{5-6}{4\sqrt{3}-0}(x-0)\Rightarrow x+4\sqrt{3}y-24\sqrt{3}=0\Rightarrow x+4\sqrt{3}y=24\sqrt{3}$

Focal radii $PS=\sqrt{{(4\sqrt{3}-0)}^2+{(5-6)}^2}=\sqrt{48+1}=7$

Equation of $P{S}^{\prime }$

$y-(-6)=\frac{5-(-6)}{4\sqrt{3}-0}(x-0)4\sqrt{3}y+24\sqrt{3}=11x11x-4\sqrt{3}y-24\sqrt{3}=0$

Focal radii $=\sqrt{{(4\sqrt{3}-0)}^2+{(5-(-6\left)\right)}^2}=\sqrt{48+121}=13$