The equation of the parabola whose focus is ( | vedclass

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Question

(A) Let $P(x, y)$ be any point on the parabola. By definition,the distance from $P$ to the focus $S(5, 3)$ is equal to the perpendicular distance from

Vedclass · Accepted Answer

$(4x + 3y)^2 - 256x - 142y + 849 = 0$

Vedclass · Answer

(A) Let $P(x, y)$ be any point on the parabola. By definition,the distance from $P$ to the focus $S(5, 3)$ is equal to the perpendicular distance from $P$ to the directrix $3x - 4y + 1 = 0$.$PS^2 = PM^2$$(x - 5)^2 + (y - 3)^2 = \left( \frac{3x - 4y + 1}{\sqrt{3^2 + (-4)^2}} \right)^2$$(x^2 - 10x + 25) + (y^2 - 6y + 9) = \frac{(3x - 4y + 1)^2}{25}$$25(x^2 + y^2 - 10x - 6y + 34) = 9x^2 + 16y^2 + 1 - 24xy + 6x - 8y$$25x^2 + 25y^2 - 250x - 150y + 850 = 9x^2 + 16y^2 - 24xy + 6x - 8y + 1$$16x^2 + 24xy + 9y^2 - 256x - 142y + 849 = 0$$(4x + 3y)^2 - 256x - 142y + 849 = 0$

Column-$I$	Column-$II$
$(A)$ Circumradius of $\Delta PQR$	$(P)$ $5/2$
$(B)$ Area of $\Delta PQR$	$(Q)$ $(5/2, 0)$
$(C)$ Centroid of $\Delta PQR$	$(R)$ $(2/3, 0)$
$(D)$ Circumcenter of $\Delta PQR$	$(S)$ $2$

The equation of the parabola whose focus is $(5, 3)$ and directrix is $3x - 4y + 1 = 0$ is:

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