If P equiv (x,;y), {F_1} equiv (3,;0), {F_2} | vedclass

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Question

(c)Equation of the curve is $\frac{{{x^2}}}{{{5^2}}} + \frac{{{y^2}}}{{{4^2}}} = 1$

$&rArr;$  $= - 5 \le x \le 5,\,\, - 4 \le y \le 4$

$P{F

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(c)Equation of the curve is $\frac{{{x^2}}}{{{5^2}}} + \frac{{{y^2}}}{{{4^2}}} = 1$

$&rArr;$  $= - 5 \le x \le 5,\,\, - 4 \le y \le 4$

$P{F_1} + P{F_2} = \sqrt {[{{(x - 3)}^2} + {y^2}]} + \sqrt {[{{(x + 3)}^2} + {y^2}]} $

$ = \sqrt {{{(x - 3)}^2} + \frac{{400 - 16{x^2}}}{{25}}} + \sqrt {{{(x + 3)}^2} + \frac{{400 - 16{x^2}}}{{25}}} $

$ = \frac{1}{5}\left\{ {\sqrt {(9{x^2} + 625 - 150x)} + \sqrt {(9{x^2} + 625 + 150x)} } \right\}$

$ = \frac{1}{5}\left\{ {\sqrt {{{(3x - 25)}^2}} + \sqrt {{{(3x + 25)}^2}} } \right\}$

$= \frac{1}{5}\left\{ {25 - 3x + 3x + 25} \right\}$

$ = 10$, $(\because 25 - 3x > 0,\,25 + 3x > 0$)

If $P \equiv (x,\;y)$, ${F_1} \equiv (3,\;0)$, ${F_2} \equiv ( - 3,\;0)$ and $16{x^2} + 25{y^2} = 400$, then $P{F_1} + P{F_2}$ equals

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