Question

Find the standard equation of the parabola that satisfies the given conditions. Also, find the length of the latus rectum of each parabola.
focus: (-3,0), directrix: x = 6
Choose the correct standard equation below.
OA.
y2 = - 18(x-3)
OB. x2 = 12(y + 3)
Ос.
x2 =
= -18
OD. y2 = 12(x+3)
Find the length of the latus rectum.
(Simplify your answer.)

Answer 1

Answer:

The standard parabola

                                y² = -18 x +27

Length of Latus rectum = 4 a = 18

                         

Step-by-step explanation:

Explanation:-

Given focus : (-3 ,0) ,directrix  : x=6

Let P(x₁ , y₁) be the point on parabola

PM perpendicular to the the directrix L

                          SP² = PM²

                (x₁ +3)²+(y₁-0)²  = $(\frac{x_{1}-6 }{\sqrt{1} } )^{2}$

              x₁²+6 x₁ +9 + y₁² = x₁²-12 x₁ +36

                          y₁² = -18 x₁ +36 -9

                           y₁² = -18 x₁ +27

The standard parabola

                                y² = -18 x +27

    Length of Latus rectum = 4 a = 4 (18/4) = 18