Equation of the parabola, if coordinates of vertex and focus are $(0, 0)$ and $(2, 3)$ respectively, is

4 x^{2} + 9 y^{2} - 156 x - 104 y - 12 x y = 0

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9 x^{2} + 4 y^{2} - 156 x - 104 y - 12 x y = 0

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9 x^{2} + 4 y^{2} - 104 x - 156 y - 12 x y = 0

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9 x^{2} + 4 y^{2} - 104 x + 156 y + 12 x y = 0

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Solution

The correct option is C $9 x^{2} + 4 y^{2} - 104 x - 156 y - 12 x y = 0$
Given, Vertex $\equiv (0, 0),$ Focus $\equiv (2, 3)$
Length of latus rectum $(L . L . R .) = 4 \sqrt{(2 - 0)^{2} + (3 - 0)^{2}} = 4 \sqrt{13}$

Equation of the axis of parabola is
$y - 0 = \frac{3 - 0}{2 - 0} (x - 0) \Rightarrow 3 x - 2 y = 0$

Equation of the tangent at vertex is
$y - 0 = \frac{- 2}{3} (x - 0) \Rightarrow 2 x + 3 y = 0$

Let $(h, k)$ be any point on the parabola.
Then equation of the parabola is
${(Distance from axis)}^{2} = L . L . R . (distance from T.V.)$
$\Rightarrow {(\frac{| 3 h - 2 k |}{\sqrt{3^{2} + 2^{2}}})}^{2} = 4 \sqrt{13} (\frac{2 h + 3 k}{\sqrt{2^{2} + 3^{2}}})$
$\Rightarrow (3 h - 2 k)^{2} = 52 (2 h + 3 k)$
$\Rightarrow 9 h^{2} + 4 k^{2} - 104 h - 156 k - 12 h k = 0$
Hence, required equation is $9 x^{2} + 4 y^{2} - 104 x - 156 y - 12 x y = 0$

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Similar questions

If the coordinates of the vertex and the focus of a parabola are (-1,1) and (2,3) respectively,then the equation of its directrix is

(- 1, 1)

(2, 3)

(1, 2)

(4, 6)

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