If P-x- y- F-1-3- 0- F-2-3- 0- and 16x-2-25y-2-400- then PF-1-PF-2 equals- 8-12-6-10-

The correct option is D 10 Given, 16x^2+25y^2=400 is the equation of ellipse. ⇒x^2/25+y^2/16=1 Here, a^2=25, b^2=16 We know, that in an ellipse distance of any ...

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Standard XII

Mathematics

Major Axis of Ellipse

If P=(x, y), ...

Question

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

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Solution

The correct option is C
10

Given, $16 x^{2} + 25 y^{2} = 400$ is the equation of ellipse.
$\Rightarrow \frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$
Here, $a^{2} = 25, b^{2} = 16$

We know, that in an ellipse distance of any point from its foci is constant and equal to length of the major axis, i.e.

$P F_{1} + P F_{2}$ = 2a

Now, $P F_{1} + P F_{2}$ = Major axis = 2a
$= 2 \times 5 = 10$

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Similar questions

Q. Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
(a) 16x² − 9y² = 144
(b) 9x² − 16y² = 144
(c) 25x² − 9y² = 225
(d) 9x² − 25y² = 81

Q. If

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

and

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

, then

P F_{1} + P F_{2}

equals

Q. If

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

and

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

, then

P F_{1} + P F_{2}

equals

Q. If

F_{1} = (3, 0)

F_{2} = (- 3, 0)

and

P

is any point on the curve

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

, then

P F_{1} + P F_{2}

equals to:

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

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If P=(x,y),F1=(3,0),F2=(−3,0) and 16x2+25y2=400, then PF1+PF2 equals

The correct option is C 10 Given, 16x2+25y2=400 is the equation of ellipse. ⇒x225+y216=1 Here, a2=25,b2=16 We know, that in an ellipse distance of any point from its foci is constant and equal to length of the major axis, i.e. PF1+PF2 = 2a Now, PF1+PF2 = Major axis = 2a =2×5=10

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