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10-2. Parabola, Ellipse, Hyperbola
easy
यदि $P \equiv (x,\;y)$, ${F_1} \equiv (3,\;0)$, ${F_2} \equiv ( - 3,\;0)$ और $16{x^2} + 25{y^2} = 400$ तो $P{F_1} + P{F_2}$ का मान है
A
$8$
B
$6$
C
$10$
D
$12$
(IIT-1998)
Solution
(c) दिये गये वक्र का समीकरण $\frac{{{x^2}}}{{{5^2}}} + \frac{{{y^2}}}{{{4^2}}} = 1$ है।
$ – 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$
Standard 11
Mathematics
