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Q.
The sum of the focal distances of any point on the conic $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ is
Bihar CECEBihar CECE 2010
Solution:
Comparing the given equation,
$\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$
with general equation of ellipse
$\frac{x^{2}}{-a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$a^{2}=25$
$ \Rightarrow a=5$
We know that, the sum of focal distances $=2\, a $
$=2 \times 5 =10$