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જો $P \equiv (x,\;y)$, ${F_1} \equiv (3,\;0)$, ${F_2} \equiv ( - 3,\;0)$ અને $16{x^2} + 25{y^2} = 400$, તો $ P{F_1} + P{F_2}$ = .. . . .
$8$
$6$
$10$
$12$
Solution
(c)Equation of the curve is $\frac{{{x^2}}}{{{5^2}}} + \frac{{{y^2}}}{{{4^2}}} = 1$
$⇒$ $= – 5 \le x \le 5,\,\, – 4 \le y \le 4$
$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$, $(\because 25 – 3x > 0,\,25 + 3x > 0$)
