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(D) The length of the latus rectum $4a$ is the distance from the point $(2, -3)$ to the line $3x + 4y - 5 = 0$.$4a = \frac{|3(2) + 4(-3) - 5|}{\sqrt{3

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$11 \frac{d^{2}y}{dx^{2}} = 10$

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(D) The length of the latus rectum $4a$ is the distance from the point $(2, -3)$ to the line $3x + 4y - 5 = 0$.$4a = \frac{|3(2) + 4(-3) - 5|}{\sqrt{3^{2} + 4^{2}}} = \frac{|6 - 12 - 5|}{5} = \frac{|-11|}{5} = \frac{11}{5}$.Since the axis is parallel to the $y$-axis,the equation of the parabola is $(x - h)^{2} = 4a(y - k)$,where $4a = \frac{11}{5}$.$(x - h)^{2} = \frac{11}{5}(y - k)$.Differentiating with respect to $x$:$2(x - h) = \frac{11}{5} \frac{dy}{dx}$.Differentiating again with respect to $x$:$2 = \frac{11}{5} \frac{d^{2}y}{dx^{2}}$.$10 = 11 \frac{d^{2}y}{dx^{2}}$,which is $11 \frac{d^{2}y}{dx^{2}} = 10$.

$A$ differential equation representing the family of parabolas with axis parallel to the $y$-axis and whose length of latus rectum is the distance of the point $(2, -3)$ from the line $3x + 4y = 5$,is given by:

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