The point (3,4) is the focus and 2x - 3y + 5 | vedclass

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(A) The focus of the parabola is $(3,4)$ and the equation of the directrix is $2x - 3y + 5 = 0$. The distance $d$ from the focus $(x_1, y_1)$ to the l

Vedclass · Accepted Answer

$\frac{2}{\sqrt{13}}$

Vedclass · Answer

(A) The focus of the parabola is $(3,4)$ and the equation of the directrix is $2x - 3y + 5 = 0$. The distance $d$ from the focus $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Here,$d = \frac{|2(3) - 3(4) + 5|}{\sqrt{2^2 + (-3)^2}} = \frac{|6 - 12 + 5|}{\sqrt{4 + 9}} = \frac{|-1|}{\sqrt{13}} = \frac{1}{\sqrt{13}}$. The length of the latus rectum of a parabola is $2 	imes$ (distance from the focus to the directrix). Therefore,the length of the latus rectum $= 2 	imes \frac{1}{\sqrt{13}} = \frac{2}{\sqrt{13}}$.

The point $(3,4)$ is the focus and $2x - 3y + 5 = 0$ is the directrix of a parabola. Its latus rectum is:

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