In an ellipse with its centre at the origin,i | vedclass

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(B) Given that the focus is at $(0, 5\sqrt{3})$,the ellipse is vertical with its major axis along the $y$-axis. Thus,$b > a$.The focus is $(0, be) = (

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$5$

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(B) Given that the focus is at $(0, 5\sqrt{3})$,the ellipse is vertical with its major axis along the $y$-axis. Thus,$b > a$.The focus is $(0, be) = (0, 5\sqrt{3})$,so $be = 5\sqrt{3}$.Squaring both sides,$b^2e^2 = 75$.For an ellipse,$a^2 = b^2(1 - e^2) = b^2 - b^2e^2$,so $b^2e^2 = b^2 - a^2 = 75$.We are given the difference between the lengths of the major axis $(2b)$ and the minor axis $(2a)$ is $10$:$2b - 2a = 10 \Rightarrow b - a = 5$.Using the identity $b^2 - a^2 = (b - a)(b + a) = 75$:$5(b + a) = 75 \Rightarrow b + a = 15$.Solving the system:$b - a = 5$$b + a = 15$Adding these gives $2b = 20 \Rightarrow b = 10$.Subtracting gives $2a = 10 \Rightarrow a = 5$.The length of the latus rectum for a vertical ellipse is $\frac{2a^2}{b}$.$LR = \frac{2(5^2)}{10} = \frac{2 	imes 25}{10} = \frac{50}{10} = 5$.

In an ellipse with its centre at the origin,if the difference between the lengths of the major axis and the minor axis is $10$ and one of the foci is at $(0, 5\sqrt{3})$,then the length of its latus rectum is:

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