If coordinates of vertex and focus of a parabola are $(1, 2)$ and $(4, 6)$ respectively, then

equation of the parabola is

(4 x - 3 y + 2)^{2} = 100 (3 x + 4 y - 11)

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equation of the parabola is

(3 x + 4 y - 11)^{2} = 100 (4 x - 3 y + 2)

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length of latus rectum of the parabola is

5

units

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length of latus rectum of the parabola is

20

units

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Solution

The correct options are
A equation of the parabola is $(4 x - 3 y + 2)^{2} = 100 (3 x + 4 y - 11)$
D length of latus rectum of the parabola is $20$ units
Given, Vertex $\equiv (1, 2),$ Focus $\equiv (4, 6)$
Length of latus rectum $(L . L . R .) = 4 \sqrt{(6 - 2)^{2} + (4 - 1)^{2}} = 20$

Equation of the axis of parabola is
$y - 2 = \frac{6 - 2}{4 - 1} (x - 1) \Rightarrow 4 x - 3 y + 2 = 0$

Equation of the tangent at vertex is
$y - 2 = \frac{- 3}{4} (x - 1) \Rightarrow 3 x + 4 y - 11 = 0$

Let $(h, k)$ be any point on the parabola.
Then equation of the parabola is
${(Distance from axis)}^{2} = L . L . R . (distance from T.V.)$
$\Rightarrow {(\frac{| 4 h - 3 k + 2 |}{\sqrt{4^{2} + 3^{2}}})}^{2} = 20 (\frac{| 3 h + 4 k - 11 |}{\sqrt{3^{2} + 4^{2}}})$
$\Rightarrow (4 h - 3 k + 2)^{2} = 100 (3 h + 4 k - 11)$

Hence, required equation is $(4 x - 3 y + 2)^{2} = 100 (3 x + 4 y - 11)$

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