Vertex is $(4, 3)$ and directrix is $3 x + 2 y - 7 = 0$ then equation of latusrectum is

3 x + 2 y - \frac{21 + 7 \sqrt{3}}{\sqrt{13}} = 0

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3 x + 2 y - \frac{22 + 7 \sqrt{3}}{\sqrt{13}} = 0

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3 x + 2 y - \frac{21 + 7 \sqrt{3}}{\sqrt{16}} = 0

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3 x + 2 y - \frac{22 + 7 \sqrt{3}}{\sqrt{15}} = 0

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Solution

The correct option is B $3 x + 2 y - \frac{22 + 7 \sqrt{3}}{\sqrt{13}} = 0$
Vertex $(4, 3)$
Directrix $3 x + 2 y - 7 = 0 . . . . . . . . . . . . . . . . (1)$
Equation of $L R$ is a line parallel to directrix and distance between them is $2 \times ($ distance of vertex from directrix $)$
Equation will be, $3 x + 2 y + c = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . (2) (c =$ constant $)$
Distance between $(4, 3)$ and $3 x + 2 y - 7 = 0$
$=> A V = ∣ ∣ ∣ \frac{3 (4) + 2 (3) - 7}{\sqrt{9 + 4}} ∣ ∣ ∣$
$=> A V = \frac{11}{\sqrt{13}}$
distance between equation $(1)$ and $(2) => 2 (A V) = \frac{22}{\sqrt{13}}$
$=> c = - ∣ ∣ ∣ - 7 - \frac{22}{\sqrt{13}} ∣ ∣ ∣ = - \frac{22 + 7 \sqrt{3}}{\sqrt{13}}$
Equation of Latus Rectum becomes, $3 x + 2 y - \frac{22 + 7 \sqrt{3}}{\sqrt{13}} = 0$

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