Which is the equation of a parabola with vertex (0, 0), that opens to the left and has a focal width of 12?

Question

asked Oct 27, 2018 40.7k views

2 Answers

Akxaya · Answer 1 · 2018-10-31T13:18:41+0000

Answer:

$\text{Equation of parabola is }y^2=-12x$

Explanation:

We need to find the equation of parabola using given information

If parabola open left and passes through origin then equation is

y^2=-4ax

Focal width = 12

Focal width passes through focus and focus is mid point of focal width.

Focus of above parabola would be (-a,0)

Passing point on parabola (-a,6) and (-a,-6)

Now we put passing point into equation and solve for a

6^2=-4a(-a)

$a=\pm 3$

a can't be negative.

Therefore, a=3

Focus: (-3,0)

Equation of parabola:

y^2=-12x

Please see the attachment of parabola.

$\text{Thus, Equation of parabola is }y^2=-12x$