The equation of an ellipse in its standard fo | vedclass

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Question

(C) The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.Given the distance between foci is $2ae = 2$,so $ae = 1$.

Vedclass · Accepted Answer

$15 x^2+16 y^2=240$

Vedclass · Answer

(C) The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.Given the distance between foci is $2ae = 2$,so $ae = 1$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{15}{2}$,which implies $b^2 = \frac{15a}{4}$.Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$,we substitute $ae = 1$ and $b^2 = \frac{15a}{4}$:$\frac{15a}{4} = a^2 - 1$$4a^2 - 15a - 4 = 0$$(4a + 1)(a - 4) = 0$Since $a$ must be positive,$a = 4$.Then $b^2 = \frac{15(4)}{4} = 15$.Substituting $a^2 = 16$ and $b^2 = 15$ into the standard equation:$\frac{x^2}{16} + \frac{y^2}{15} = 1$$15x^2 + 16y^2 = 240$.

The equation of an ellipse in its standard form,given the distance between its foci is $2$ units and the length of its latus rectum is $\frac{15}{2}$ units,is

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