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If the focus and vertex of a parabola are at (0,2) and (0,4) respectively, then the equation of the parabola is
Option: 1
$\mathrm{x^2+8 y-32=0}$

Option: 2
$\mathrm{x^2-8 y+32=0}$

Option: 3
$\mathrm{y^2=-8 x+32}$

Option: 4
$\mathrm{x^2-8 y-32=0}$

Answers (1)

best_answer

As vertex (0,4) and focus (0,2) lie on the y-axis and the vertex is above the focus, the curve lies on the third and fourth quadrants.

Distance between the vertex and the focus $\mathrm{=4-2=2 \equiv a}$

$\mathrm{\therefore \quad}$ The equation of the directrix
$\mathrm{\text { is } y=(y \text {-coordinate of vertex })+\frac{1}{4}(\mathrm{~L} \cdot \mathrm{R})}$

$\mathrm{\begin{aligned} & =4+a \\ & =4+2=6 \end{aligned}}$

The directrix is $\mathrm{y-6=0}$

And, the focus is $\mathrm{(0,2).}$

If $\mathrm{P\left(x_1, y_1\right)}$ is a point on the parabola, then by definition of a parabola,

$\mathrm{ \sqrt{\left(x_1-0\right)^2+\left(y_1-2\right)^2}=\frac{y_1-6}{\sqrt{1}} }$
Or $\mathrm{\quad x_1^2+\left(y_1-2\right)^2=\left(y_1-6\right)^2}$

$\mathrm{\therefore \quad \text { The locus of } P\left(x_1, y_1\right) \text { is }}$

$\mathrm{ x^2+(y-2)^2=(y-6)^2 }$

$\mathrm{\text { Or } \quad x^2=-8 y+32}$

The answer is (a)

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Pankaj

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