The length of the latus rectum of 16x^2 + 25y | vedclass

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(D) Given the equation of the ellipse: $16x^2 + 25y^2 = 400$. Dividing both sides by $400$,we get: $\frac{16x^2}{400} + \frac{25y^2}{400} = 1$,which s

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$\frac{32}{5}$

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(D) Given the equation of the ellipse: $16x^2 + 25y^2 = 400$. Dividing both sides by $400$,we get: $\frac{16x^2}{400} + \frac{25y^2}{400} = 1$,which simplifies to $\frac{x^2}{25} + \frac{y^2}{16} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 16$. Thus,$a = 5$ and $b = 4$. The length of the latus rectum is given by the formula $\frac{2b^2}{a}$. Substituting the values: $\frac{2 	imes 16}{5} = \frac{32}{5}$.

The length of the latus rectum of $16x^2 + 25y^2 = 400$ is

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