Find the equation of the parabola whose focus is the point (2,3) and directrix is the line x-4y+3=0.Also,find the length of its latus-rectum.

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Solution

Let P(x,y) be any point on the parabola whose focus is S(2,3) and the directirx x-4y+3=0.

Draw PM perpendicular from P(x,y) on the directrix x-4y+3=0.

Then by definition SP=PM

$\Rightarrow S P^{2} = P M^{2} \Rightarrow (x - 2)^{2} + (y - 3)^{2} = {(\frac{x - 4 y + 3}{\sqrt{(1)^{2} + (- 4)^{2}}})}^{2} \Rightarrow x^{2} + 4 - 4 x + y^{2} + 9 - 6 y = (\frac{(x - 4 y + 3)^{2}}{\sqrt{(17)^{2})}}) \Rightarrow x^{2} + y^{2} - 4 x - 6 y + 4 + 9 = \frac{(x - 4 y + 3)^{2}}{17} \Rightarrow 17 (x^{2} + y^{2} - 4 x - 6 y + 13) = (x - 4 y + 3)^{2} \Rightarrow 17 x^{2} + 17 y^{2} - 68 x - 102 y + 221 = x^{2} + (- 4 y)^{2} + 3^{2} + 2 \times 3 \times x \Rightarrow 17 x^{2} + 17 y^{2} - 68 x - 102 y + 221 = x^{2} + 16 y^{2} + 9 - 8 x y - 24 y + 6 x \Rightarrow 17 x^{2} - x^{2} + 17 y^{2} - 16 y^{2} + 8 x y - 68 x - 6 x - 102 y + 24 y + 221 - 9 = 0 \Rightarrow 16 x^{2} + y^{2} + 8 x y - 74 x - 78 y + 212 = 0$

This is the equation of the required parabola.

Latus Rectum=Length of perpendicular from focus (2,3) on directrix x-4y+3=0

$= 2 ∣ ∣ ∣ \frac{2 - 12 + 3}{\sqrt{1 + 16}} ∣ ∣ ∣ = 2 ∣ ∣ ∣ \frac{- 7}{\sqrt{17}} ∣ ∣ ∣ = \frac{14}{\sqrt{17}}$

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Similar questions

(2, 3)

x - 4 y + 3 = 0.

(- 2, 3)

2 x + 3 y - 4 = 0

Length of latus rectum of the parabola whose focus is at $(2, 3)$ and directrix is the line $x - 4 y + 3 = 0$ is

(3, 0)

x = - 3

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