Let the length of the latus rectum of an elli | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The length of the latus rectum is $\frac{2b^2}{a} = 8$,which

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$(4\sqrt{3}, 2\sqrt{2})$

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(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The length of the latus rectum is $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.The distance between the foci is $2ae$ and the length of the minor axis is $2b$. Given $2ae = 2b$,we have $ae = b$,so $a^2e^2 = b^2$.Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$,we substitute $a^2e^2 = b^2$ to get $b^2 = a^2 - b^2$,or $2b^2 = a^2$.Substituting $b^2 = 4a$ into $2b^2 = a^2$,we get $2(4a) = a^2$,so $a^2 = 8a$. Since $a 
eq 0$,$a = 8$.Then $b^2 = 4(8) = 32$,so $b = \sqrt{32} = 4\sqrt{2}$.The equation of the ellipse is $\frac{x^2}{64} + \frac{y^2}{32} = 1$.Testing the points,for $(4\sqrt{3}, 2\sqrt{2})$: $\frac{(4\sqrt{3})^2}{64} + \frac{(2\sqrt{2})^2}{32} = \frac{48}{64} + \frac{8}{32} = \frac{3}{4} + \frac{1}{4} = 1$. Thus,the point lies on the ellipse.

Let the length of the latus rectum of an ellipse with its major axis along the $x$-axis and center at the origin be $8$. If the distance between the foci of this ellipse is equal to the length of its minor axis,then which one of the following points lies on it?

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