The equation of the directrix of the parabola | vedclass

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Question

(B) The given equation of the parabola is $(2 x - 3 y - 5)^2 = 20(3 x + 2 y + 1)$.
We rewrite this in the form $\left( \frac{2 x - 3 y - 5}{\sqrt{2^2

Vedclass · Accepted Answer

$3 x + 2 y + 1 - 5 = 0$

Vedclass · Answer

(B) The given equation of the parabola is $(2 x - 3 y - 5)^2 = 20(3 x + 2 y + 1)$.
We rewrite this in the form $\left( \frac{2 x - 3 y - 5}{\sqrt{2^2 + (-3)^2}} \right)^2 = \frac{20}{13} \left( \frac{3 x + 2 y + 1}{\sqrt{3^2 + 2^2}} \right) \sqrt{13}$.
Let $Y = \frac{2 x - 3 y - 5}{\sqrt{13}}$ and $X = \frac{3 x + 2 y + 1}{\sqrt{13}}$.
The equation becomes $Y^2 = \frac{20}{\sqrt{13}} X$.
Comparing with $Y^2 = 4 a X$,we get $4 a = \frac{20}{\sqrt{13}}$,so $a = \frac{5}{\sqrt{13}}$.
The directrix is given by $X = -a$,which is $\frac{3 x + 2 y + 1}{\sqrt{13}} = -\frac{5}{\sqrt{13}}$.
Thus,$3 x + 2 y + 1 = -5$,or $3 x + 2 y + 6 = 0$.

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

The equation of the directrix of the parabola $(2 x - 3 y - 5)^2 = 20(3 x + 2 y + 1)$ is

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