If F_1 and F_2 are the foci of the ellipse 16 | vedclass

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(A) Given,equation of ellipse $16 x^2+25 y^2=400$ $\Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1$. Comparing to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we g

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$[16, 25]$

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(A) Given,equation of ellipse $16 x^2+25 y^2=400$ $\Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1$. Comparing to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we get $a=5$ and $b=4$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. Let $P(5 \cos 	heta, 4 \sin 	heta)$ be any point on the ellipse. The focal distances are $P F_1 = a - ex = 5 - (\frac{3}{5})(5 \cos 	heta) = 5 - 3 \cos 	heta$ and $P F_2 = a + ex = 5 + 3 \cos 	heta$. Now,the product $P F_1 \cdot P F_2 = (5 - 3 \cos 	heta)(5 + 3 \cos 	heta) = 25 - 9 \cos^2 	heta$. Since $0 \leq \cos^2 	heta \leq 1$,we have: $25 - 9(1) \leq 25 - 9 \cos^2 	heta \leq 25 - 9(0)$ $16 \leq P F_1 \cdot P F_2 \leq 25$. Thus,the value lies in the interval $[16, 25]$.

If $F_1$ and $F_2$ are the foci of the ellipse $16 x^2+25 y^2=400$ and $P$ is any point on it,then the value of the product $P F_1 \cdot P F_2$ lies in the interval

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