Let an ellipse with centre (1,0) and latus re | vedclass

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

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$9$

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(A) The equation of the ellipse is $\frac{(x-1)^2}{a^2} + \frac{y^2}{b^2} = 1$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{1}{2}$,which implies $4b^2 = a$ (Equation $1$).The minor axis endpoints are $(1, b)$ and $(1, -b)$,and the foci are $(1 \pm ae, 0)$. The angle subtended by the minor axis at a focus is $60^{\circ}$.Considering the triangle formed by the focus $(1+ae, 0)$ and the endpoints of the minor axis $(1, b)$ and $(1, -b)$,the angle at the focus is $60^{\circ}$,so the half-angle is $30^{\circ}$.Thus,$	an(30^{\circ}) = \frac{b}{ae} = \frac{1}{\sqrt{3}}$,which implies $ae = b\sqrt{3}$.Squaring both sides,$a^2e^2 = 3b^2$. Since $a^2e^2 = a^2 - b^2$,we have $a^2 - b^2 = 3b^2$,so $a^2 = 4b^2$ (Equation $2$).From Equation $1$,$b^2 = \frac{a}{4}$. Substituting this into Equation $2$,$a^2 = 4(\frac{a}{4}) = a$.Since $a > 0$,we get $a = 1$. Then $b^2 = \frac{1}{4}$,so $b = \frac{1}{2}$.The length of the major axis is $2a = 2(1) = 2$,and the length of the minor axis is $2b = 2(\frac{1}{2}) = 1$.The square of the sum of the lengths is $(2a + 2b)^2 = (2 + 1)^2 = 3^2 = 9$.

Let an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$-axis. If its minor axis subtends an angle $60^{\circ}$ at the foci,then the square of the sum of the lengths of its minor and major axes is equal to $...........$.

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