The focal distances of the point 4 5 , 3 5 on the ellipse x 2 4 + y 2 9 = 1 are

Given,

Equation of ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$,

So we can see $b>a$ so it is vertical ellipse,

Hence eccentricity is given by $a^2=b^2\left(1-e^2\right)$,

$\Rightarrow 4=9\left(1-e^2\right)$

$\Rightarrow e=\frac{\sqrt{5}}{3}$

Now focus is given by $\left(0,\pm be\right)\equiv \left(0,\pm \sqrt{5}\right)$

Now assuming $S\equiv \left(0,\sqrt{5}\right) \& S^{\text{'}}\equiv \left(0,-\sqrt{5}\right)$ and point $P\left(\frac{4}{\sqrt{5}},\frac{3}{\sqrt{5}}\right)$,

Now focal distance will be $PS \& P{S}^{\text{'}}$,

Now using distance formula we get, $PS=\sqrt{{\left(\frac{4}{\sqrt{5}}-0\right)}^2+\left(\frac{3}{\sqrt{5}}-\sqrt{5}\right)}=\sqrt{4}=2$

And similarly $P{S}^{\text{'}}=4$