$A (0, 1)$ and $B (0, - 1)$ are $2$ points. If a variable point P moves such that sum of its distance fro A and B is $4$ . Then the locus of P is the equation of the form of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ . Find the value of $(a^{2} + b^{2})$ .

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Solution

$A (0, 1)$ & $B (0, - 1)$ are the focii of required ellipse.
Also distance between focii = 2 ae
$∴$ 2 ae = 2
ae = 1
also PA + PB = 4
According to definition of ellipse,
$P F_{1} + P F_{2} = 2 a$
$∴ 2 a = 4$
$\Rightarrow a = 2$
$\Rightarrow e = \frac{1}{2}$
$\sqrt{1 - \frac{b^{2}}{a^{2}}} = \frac{1}{2} \Rightarrow b^{2} = 3$
$∴ a^{2} + b^{2} = (2)^{2} + 3$
$= 4 + 3$
$= 7$

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