Question

The focus of a parabola is $$\left(1,5\right)$$and its directrix is$$x+y+2=0.$$Find the equation of the parabola. Its vertex and length of latus rectum?

Answer 1

Hint: First by definition of parabola we find the equation of parabola then find the equation of axis using the slope of directrix which will give the value of coordinates then using formula Length of latus rectum $$=4\times |AF|$$. we get the answer.

Complete step by step solution:
Given: The focus of a parabola is $$\left(1,5\right)$$ and the equation of directrix is $$x+y+2=0\dots \left[1\right]$$
Let $$P\left(h,k\right)$$ be any point on the parabola.
Then by definition of parabola, we have
The distance from the focus to the point P = the distance from directrix to the point P
i.e. $$|PF|=|PM|$$
$$\Rightarrow \sqrt{{\left(h-1\right)}^2+{\left(k-5\right)}^2}={\displaystyle \frac{|h+k+2|}{\sqrt{1+1}}}$$
Squaring on both sides we get,
$$\Rightarrow {\left(h-1\right)}^2+{\left(k-5\right)}^2={\displaystyle \frac{{\left(h+k+2\right)}^2}{2}}$$
$$\Rightarrow 2\times \left(h^2-2h+1+k^2-10k+25\right)=\left(h^2+k^2+4+2hk+4h+4k\right)$$
$$\Rightarrow$$$$2h^2-4h+2+2k^2-20k+50=h^2+k^2+4+2hk+4h+4k$$
$$\Rightarrow h^2+k^2-8h-24k-2hk+48=0\dots \left[2\right]$$
Put $$h=x$$ and $$k=y$$in [2]
$$\Rightarrow x^2+y^2-8x-24y-2xy+48=0\dots \left[3\right]$$is the required equation of parabola.
The axis of parabola passes through the focus and perpendicular to the directrix.
The slope of axis $$=-\left(-1\right)$$
 $$=1$$ [Since the slope of the directrix is -1]
The equation of axis is $$y-5=1\times \left(x-1\right)$$
$$\Rightarrow y-5=\left(x-1\right)$$
$$\Rightarrow$$$$x-y+4=0\dots \left[4\right]$$
The point of intersection of axis and directrix is the intersection point of equation [1] and [4]
On solving $$x+y+2=0$$and $$x-y+4=0$$we get,
$$2x+6=0$$
$$\Rightarrow x=-{\displaystyle \frac{6}{2}}$$
$$\Rightarrow x=-3$$; putting this value in $$x+y+2=0$$we get,
$$-3+y+2=0$$
$$\Rightarrow y=1$$
The coordinates of K are $$\left(-3,1\right)$$
Clearly, vertex is a midpoint of focus and point K
i.e. coordinates of vertex are $$\left({\displaystyle \frac{1-3}{2}},{\displaystyle \frac{5+1}{2}}\right)$$
$$=\left(-1,3\right)$$
Now, distance from vertex to focus $$=\sqrt{{\left(-1-1\right)}^2+{\left(3-5\right)}^2}$$
$$=\sqrt{4+4}$$
$$=2\sqrt{2}$$
Length of latus rectum$$=4\times |AF|$$
$$=4\times 2\sqrt{2}$$
$$=8\sqrt{2}$$

Note: Parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane. Fixed-line is called Directrix of the parabola and the fixed point is called Focus usually we denote Focus point as (F).