Equation of the parabola having focus $(3, 2)$ and Vertex $(- 1, 2)$ is

(x + 1)^{2} = 16 (y - 2)

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(x - 1)^{2} = 16 (y + 2)

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(y - 2)^{2} = 16 (x + 1)

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(y + 2)^{2} = 16 (x - 1)

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Solution

The correct option is C $(y - 2)^{2} = 16 (x + 1)$
Equation of a General Parabola having axis along the x-axis is given by $y^{2} = 4 a x$

where $a =$ distance between Vertex and Focus.

Here $y$ coordinate in Focus and Vertex is the same.

Thus this parabola would be along direction of the x-axis.

Diatance between vertex and focus $= \sqrt{(3 - (- 1))^{2} + (2 - 2)^{2}} = 4$

thus equation of parabola is $(y - 2)^{2} = 4. (4) (x - (- 1))$

i.e. $(y - 2)^{2} = 16 (x + 1)$

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Similar questions

8

x -

y -

y^{2} = 16 x

x^{2} = 16 y

List-l	List - II
A. Length of the latusrectum of the parabola $3 y^{2} = - 4 x$	1. $(5, 2)$
B.Equation of the parabola whose focus $($ 0,4 $)$ and directrix $x + 4 = 0$ is	2. $\frac{- 4}{3}$
C. The focus of the parabola $(y - 2)^{2} = 8 (x - 3)$ is	3. $\frac{4}{3}$
D. Equation of the parabola whose vertex $(- 1, - 2)$ latusrectum 4 and axis is parallel to y-axis	4. $(x + 1)^{2} = 4 (y + 2)$
	5. $x^{2} = 16 y$

\sqrt{(x - 1)^{2} + (y - 2)^{2}} = \sqrt{λ} | x + y + 3 |

(x - 2)^{2} = 16 (y - 4)

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