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Focal Chord of Ellipse

If P = x, y...

Question

If $P = (x, y), F_{1} = (3, 0)$ and $16 x^{2} + 25 y^{2} = 400$ , then $P F_{1} + P F_{2}$ equals

8

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10

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Solution

The correct option is B $10$
$P F_{1} + P F_{2} = 2 a$

$\frac{x^{2}}{5^{2}} + \frac{y^{2}}{4^{2}} = 1 \Rightarrow a = 5, b = 4$

$∴ P F_{1} + P F_{2} = 2 (5)$

$= 10$

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Similar questions

P = (x, y), F_{1} = (3, 0), F_{2} = (- 3, 0)

16 x^{2} + 25 y^{2} = 400

P F_{1} + P F_{2}

F_{1} = (3, 0)

F_{2} = (- 3, 0)

P

16 x^{2} + 25 y^{2} = 400

P F_{1} + P F_{2}

If P (x, y) is any point on the ellipse

16 x^{2} + 25 y^{2} = 400

and

F_{1} = (3, 0), F_{2} = (- 3, 0), then the value of {PF}_{1} + {PF}_{2} is

16 x^{2} + 25 y^{2} = 400.

P S Q

16 x^{2} + 25 y^{2} = 400

S P = 8

S Q =

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