Question

Find the equation of the parabola whose focus is at (-2,-1) and the latus rectum joins the points (-2,2) and (-2,-4

Answer 1

Answer:

The equation of parabola is $(y+1)^2=6x+21$.

Step-by-step explanation:

The focus of parabola is (-2,-1). The latus rectum joins the points (-2,2) and (-2,-4). The latus rectum is a vertical line so the parabola is along the x-axis.

Let the equation of the parabola is

$(y-k)^2=4p(x-h)$                  ..... (1)

where, (h+p,k) is focus and 4p is the length of latus rectum.

The length of latus rectum is

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{((-2+2)^2+(-4-2)^2}=6$

$4p=6$

$p=\frac{6}{4}=\frac{3}{2}$

The focus of parabola is (-2,-1).

$(h+p,k)=(-2,-1)$

$(h+\frac{3}{2},k)=(-2,-1)$

On comparing both sides we get,

$h+\frac{3}{2}=-2\Rightarrow h=-\frac{7}{2}$

$k=-1$

Put these values in equation (1).

$(y-(-1))^2=4(\frac{3}{2})(x-(-\frac{7}{2}))$

$(y+1)^2=6(x+\frac{7}{2})$

$(y+1)^2=6x+21$

Therefore the equation of parabola is $(y+1)^2=6x+21$.

Answer 2

Answer:

y²+2y-6x-20=0 and y²+6x+2y+4=0