Chapter 10.2, Problem 56PPSE

Solution

To Show: The parabola with the given vertex and the given focus has the given equation, using the definition of a parabola.

It has been shown using the definition of a parabola, that the parabola with the vertex, $\left(h,k\right)$ and the focus, $\left(h,k+c\right)$ has the equation, ${\left(x-h\right)}^2=4c\left(y-k\right)$ .

Given:

The vertex, $\left(h,k\right)$ , the focus, $\left(h,k+c\right)$ and the equation, ${\left(x-h\right)}^2=4c\left(y-k\right)$ .

Concept used:

A parabola is defined as the locus of a point which is equidistant from a point called focus and a line called the directrix.

Calculation:

As discussed, a parabola is defined as the locus of a point which is equidistant from a point called focus and a line called the directrix.

It is given that the parabola has its vertex at $\left(h,k\right)$ and its focus at $\left(h,k+c\right)$ .

Note that these points both lie on the line $x=h$ , which is the axis of symmetry.

Then, the directrix, which is perpendicular to the axis of symmetry, must be a horizontal line of the form $y=a$ .

The foot of the perpendicular drawn from the vertex, $\left(h,k\right)$ to the line, $y=a$ must be $\left(h,a\right)$ .

Since the vertex lies on the parabola, it must be equidistant from the focus and directrix.

The distance between the vertex, $\left(h,k\right)$ and the focus, $\left(h,k+c\right)$ is $c$ .

The distance between the vertex, $\left(h,k\right)$ and the directrix, which is equal to the distance between the vertex, $\left(h,k\right)$ and the point $\left(h,a\right)$ , is $k-a$ .

According to the problem, $k-a=c$ .

Simplifying, $a=k-c$ .

So, the directrix is given as $y=k-c$ .

Now, let $\left(x,y\right)$ be any point on the parabola.

According to the definition, it is equidistant from the focus and the directrix.

The distance between $\left(x,y\right)$ and the directrix, $y=k-c$ is the distance between $\left(x,y\right)$ and the foot of the perpendicular from this point to the directrix, $y=k-c$ ; $\left(x,k-c\right)$ .

Now, the distance between $\left(x,y\right)$ and $\left(x,k-c\right)$ is $y-\left(k-c\right)$ .

Applying distance formula, the distance between $\left(x,y\right)$ and the focus, $\left(h,k+c\right)$ is given as,

  $\sqrt{{\left(x-h\right)}^2+{\left(y-\left(k+c\right)\right)}^2}$

According to the problem,

  $\sqrt{{\left(x-h\right)}^2+{\left(y-\left(k+c\right)\right)}^2}=y-\left(k-c\right)$

Squaring both sides,

  ${\left(x-h\right)}^2+{\left(y-\left(k+c\right)\right)}^2={\left(y-\left(k-c\right)\right)}^2$

Simplifying,

  $\begin{array}{c}{\left(x-h\right)}^2+{\left(y-\left(k+c\right)\right)}^2={\left(y-\left(k-c\right)\right)}^2\\ {\left(x-h\right)}^2+y^2+{\left(k+c\right)}^2-2\left(k+c\right)y=y^2+{\left(k-c\right)}^2-2\left(k-c\right)y\\ {\left(x-h\right)}^2+\left[{\left(k+c\right)}^2-{\left(k-c\right)}^2\right]=2\left(k+c\right)y-2\left(k-c\right)y\\ {\left(x-h\right)}^2+4kc=2\left[\left(k+c\right)-\left(k-c\right)\right]y\end{array}$

On further simplification,

  $\begin{array}{c}{\left(x-h\right)}^2+4kc=2\left[k+c-k+c\right]y\\ {\left(x-h\right)}^2+4kc=2\left[2c\right]y\\ {\left(x-h\right)}^2+4kc=4cy\\ {\left(x-h\right)}^2=4cy-4kc\end{array}$

Finally,

  ${\left(x-h\right)}^2=4c\left(y-k\right)$

Since $\left(x,y\right)$ is assumed to be any point on the parabola, ${\left(x-h\right)}^2=4c\left(y-k\right)$ is nothing but the equation of the parabola.

This is the required proof.

Conclusion:

It has been shown using the definition of a parabola, that the parabola with the vertex, $\left(h,k\right)$ and the focus, $\left(h,k+c\right)$ has the equation, ${\left(x-h\right)}^2=4c\left(y-k\right)$ .