Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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)

Posted by

Pankaj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2025: BTech in AI gains popularity; IIT, NIT cut-offs for artificial intelligence

anam-khan

JEE Main 2025 Live: NTA JEE session 1 begins tomorrow; admit card link, exam pattern, passing marks

anam-khan

JEE Main 2025 exam centre changed at UAE's Sharjah due to “unavoidable circumstances”

Latest Question