a and b are the semi-major and semi-minor axe | vedclass

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(A) The length of the latus rectum is given by $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$. The distance between the foci is $2ae = 4\sqrt{2}$,so $a

Vedclass · Accepted Answer

$24$

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(A) The length of the latus rectum is given by $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$. The distance between the foci is $2ae = 4\sqrt{2}$,so $ae = 2\sqrt{2}$. Squaring both sides,we get $a^2e^2 = 8$. Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $a^2(1 - \frac{b^2}{a^2}) = 8$,which simplifies to $a^2 - b^2 = 8$. Substituting $b^2 = 2a$ into the equation,we get $a^2 - 2a - 8 = 0$. Factoring the quadratic,$(a - 4)(a + 2) = 0$. Since $a > 0$,we have $a = 4$. Then $b^2 = 2(4) = 8$. Finally,$a^2 + b^2 = 4^2 + 8 = 16 + 8 = 24$.

$a$ and $b$ are the semi-major and semi-minor axes of an ellipse whose axes are along the coordinate axes. If its latus rectum is of length $4$ units and the distance between its foci is $4 \sqrt{2}$,then $a^2+b^2=$

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