Question

The focus of a parabola is (1,5) and ita directrix is the straight line x+y+2=0. Find the vertex and the latus rectum of the parabola

Answer 1

Please see the attachment

Answer 2

Answer:

Vertex(-1,3) and latus rectum=$4\sqrt2$ units

Step-by-step explanation:

We are given that a parabola with focus (1,5) and its directrix is the straight line $x+y+2=0$

$y=-x-2$

Slop of directrix=-1

Slope of axis=1

Equation of axis symmetry and axis is the perpendicular line to directrix

$y=x+k$

Using formula y=mx+c

Equation of axis passing through the focus (1,5)

Then ,we get k=5-1=4

Therefore, the equation of axis

$x-y=-4$

Foot of perpendicular from the focus is the point of intersection of axis of symmetry and directrix

By solving two equation

we get x=-3 and y=1

Vertex is the mid point of of the line which is perpendicular from focus to directrix

Using midpoint formula

$x=\frac{1-3}{2},y=\frac{5+1}{2}$

The coordinates of vertex (-1,3).

Length of latus rectum is the perpendicular distance from focus to directrix

$\frac{1(1)+1(5)+2}{\sqrt{1+1}}$

Latus rectum=$\frac{8\sqrt2}{2}$=$4\sqrt2$units