You visited us 1 times! Enjoying our articles? Unlock Full Access!

Major Axis of Ellipse

If P=x, y, F1...

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$= 2 \times 5 = 10$

Suggest Corrections

Similar questions

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

(- 3, 0)

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

4

(- 3, 0)

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

P

4.

4 x + 3 y = k,

k

Join BYJU'S Learning Program