Question

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

• A. 8
• B. 6
• C. 10
• D. 12

Answer 1

Answer: (C)

10

Answer 2

We have $\mathbf{16}\mathbf{x}^{\mathbf{2}}\mathbf{+ 25}\mathbf{y}^{\mathbf{2}}\mathbf{= 400 \Rightarrow}\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{25}}\mathbf{+}\frac{\mathbf{y}^{\mathbf{2}}}{\mathbf{16}}\mathbf{= 1}$or $\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{a}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{y}^{\mathbf{2}}}{\mathbf{b}^{\mathbf{2}}}\mathbf{= 1}$, where $a^{2} = 25$and $b^{2} = 16$

This equation represents an ellipse with eccentricity given by $e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{16}{25} = \frac{9}{25}$⇒ $e = 3/5$

So, the coordinates of the foci are $( \pm ae,0)$ i.e. $(3,0)$ and $( - 3,0)$, Thus, $F_{1}$ and $F_{2}$ are the foci of the ellipse.

Since, the sum of the focal distance of a point on an ellipse is equal to its major axis, ∴ $PF_{1} + PF_{2} = 2a = 10$