If lines 3x + 2y = 10 and -3x + 2y = 10 are t | vedclass

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(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The extremities of the latus rectum are $(\pm ae, be)$ or $(\pm ae, -b

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$\frac{100}{27}$

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(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The extremities of the latus rectum are $(\pm ae, be)$ or $(\pm ae, -be)$. Given the tangents at these points,we consider the tangent at $(ae, be)$,which is $\frac{x(ae)}{a^2} + \frac{y(be)}{b^2} = 1$,simplifying to $\frac{ex}{a} + \frac{ey}{b} = 1$. Comparing this with the given line $3x + 2y = 10$,we rewrite the line as $\frac{3}{10}x + \frac{2}{10}y = 1$,or $\frac{3}{10}x + \frac{1}{5}y = 1$. Equating coefficients: $\frac{e}{a} = \frac{3}{10}$ and $\frac{e}{b} = \frac{1}{5}$. From the second equation,$b = 5e$. Using $b^2 = a^2(1 - e^2)$,we have $25e^2 = a^2(1 - e^2)$. From $\frac{e}{a} = \frac{3}{10}$,we get $a = \frac{10e}{3}$. Substituting $a^2 = \frac{100e^2}{9}$ into the equation: $25e^2 = \frac{100e^2}{9}(1 - e^2)$. Dividing by $25e^2$ (since $e 
eq 0$): $1 = \frac{4}{9}(1 - e^2) \Rightarrow 9 = 4 - 4e^2 \Rightarrow 4e^2 = -5$,which suggests the ellipse is vertical. Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b > a$. The tangent at $(ae, be)$ is $\frac{x(ae)}{a^2} + \frac{y(be)}{b^2} = 1$. Given the lines $2y = -3x + 10$ and $2y = 3x + 10$,they intersect at $(0, 5)$. Thus,$be = 5$. The slope of the tangent is $\pm \frac{3}{2}$. The tangent equation is $y = mx \pm \sqrt{a^2m^2 + b^2}$. For $y = -\frac{3}{2}x + 5$,$m = -\frac{3}{2}$ and $c = 5$. $5^2 = a^2(-\frac{3}{2})^2 + b^2 \Rightarrow 25 = \frac{9}{4}a^2 + b^2$. Also,the point of tangency $(x_1, y_1)$ satisfies $\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$ and $y_1 = be$. Using the condition for the latus rectum,the length of the latus rectum is $\frac{2a^2}{b} = \frac{100}{27}$.

If lines $3x + 2y = 10$ and $-3x + 2y = 10$ are tangents at the extremities of the latus rectum of an ellipse whose centre is the origin,then the length of the latus rectum of the ellipse is:

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