Chapter 8.1, Problem 75E

Solution

(a.)

To Prove: The $x$ coordinates of the end points of a focal chord of the given parabola, are $x=2p\left(m\pm \sqrt{m^2+1}\right)$ , where $m$ is the slope of the focal chord.

It has been shown that the $x$ coordinates of the end points of a focal chord of the given parabola, are $x=2p\left(m\pm \sqrt{m^2+1}\right)$ , where $m$ is the slope of the focal chord.

Given:

The parabola, $x^2=4py$ .

Concept used:

The focus of a parabola $x^2=4py$ is at $\left(0,p\right)$ .

Calculation:

The given parabola is $x^2=4py$ .

Let $m$ be the slope of a focal chord.

Obviously, the focal chord passes through the focus, which is at $\left(0,p\right)$ .

Then, using the point-slope form, the equation of the straight line which the focal chord is a part of, is:

  $y-p=mx$

Simplifying,

  $y=mx+p$

Now, the end points of the focal chord are precisely the intersection points of the above straight line and the given parabola; $x^2=4py$ .

Put $y=mx+p$ in $x^2=4py$ to get,

  $x^2=4p\left(mx+p\right)$

Simplifying,

  $x^2=4pmx+4p^2$

On further simplification,

  $x^2-4pmx-4p^2=0$

Applying the quadratic formula,

  $x=\frac{-\left(-4pm\right)\pm \sqrt{{\left(-4pm\right)}^2-4\left(1\right)\left(-4p^2\right)}}{2\left(1\right)}$

Simplifying,

  $x=\frac{4pm\pm \sqrt{16p^2{m}^2+16p^2}}{2}$

On further simplification,

  $x=\frac{4pm\pm \sqrt{16p^2\left(m^2+1\right)}}{2}$

Continuing simplification,

  $x=\frac{4pm\pm 4p\sqrt{m^2+1}}{2}$

Finally,

  $x=2pm\pm 2p\sqrt{m^2+1}$

Equivalently,

  $x=2p\left(m\pm \sqrt{m^2+1}\right)$

This shows that the $x$ coordinates of the end points of a focal chord of the given parabola, are $x=2p\left(m\pm \sqrt{m^2+1}\right)$ , where $m$ is the slope of the focal chord.

Conclusion:

It has been shown that the $x$ coordinates of the end points of a focal chord of the given parabola, are $x=2p\left(m\pm \sqrt{m^2+1}\right)$ , where $m$ is the slope of the focal chord.

(b.)

To Prove: The minimum length of a focal chord is the focal width $\left|4p\right|$ .

It has been shown that the minimum length of a focal chord is the focal width $\left|4p\right|$ .

Given:

The parabola, $x^2=4py$ .

Concept used:

The focus of a parabola $x^2=4py$ is at $\left(0,p\right)$ .

Calculation:

The given parabola is $x^2=4py$ .

Let $m$ be the slope of a focal chord.

As shown previously, the $x$ coordinates of the end points of a focal chord of the given parabola, are $x=2p\left(m\pm \sqrt{m^2+1}\right)$ , where $m$ is the slope of the focal chord.

Put $x=2p\left(m\pm \sqrt{m^2+1}\right)$ in $x^2=4py$ to get,

  $\begin{array}{c}y=\frac{1}{4p}{\left(2p\left(m\pm \sqrt{m^2+1}\right)\right)}^2\\ =\frac{1}{4p}\left(4p^2{\left(m\pm \sqrt{m^2+1}\right)}^2\right)\\ =p\left(m^2+\left(m^2+1\right)\pm 2m\sqrt{m^2+1}\right)\\ =p\left(2m^2+1\pm 2m\sqrt{m^2+1}\right)\end{array}$

Then, the end points of the focal chord are: $\left(2p\left(m+\sqrt{m^2+1}\right),p\left(2m^2+1+2m\sqrt{m^2+1}\right)\right)$ and $\left(2p\left(m-\sqrt{m^2+1}\right),p\left(2m^2+1-2m\sqrt{m^2+1}\right)\right)$ .

Now, using distance formula, the length of the focal chord is given as:

  $\begin{array}{c}S=\sqrt{{\left(2p\left(m+\sqrt{m^2+1}\right)-2p\left(m-\sqrt{m^2+1}\right)\right)}^2+{\left(p\left(2m^2+1+2m\sqrt{m^2+1}\right)-p\left(2m^2+1-2m\sqrt{m^2+1}\right)\right)}^2}\\ =\sqrt{{\left(2p\left(m+\sqrt{m^2+1}-m+\sqrt{m^2+1}\right)\right)}^2+{\left(p\left(2m^2+1+2m\sqrt{m^2+1}-2m^2-1+2m\sqrt{m^2+1}\right)\right)}^2}\\ =\sqrt{{\left(2p\left(2\sqrt{m^2+1}\right)\right)}^2+{\left(p\left(4m\sqrt{m^2+1}\right)\right)}^2}\\ =\sqrt{{\left(4p\sqrt{m^2+1}\right)}^2+{\left(4mp\sqrt{m^2+1}\right)}^2}\end{array}$

Simplifying,

  $\begin{array}{c}S=\sqrt{16p^2\left(m^2+1\right)+16m^2{p}^2\left(m^2+1\right)}\\ =\sqrt{16p^2\left(m^2+1\right)\left\{m^2+1\right\}}\\ =\sqrt{16p^2{\left(m^2+1\right)}^2}\\ =\left|4p\right|\left(m^2+1\right)\end{array}$

Note that only the positive square root is considered as length cannot be negative.

So, the length of the focal chord is given as $\left|4p\right|\left(m^2+1\right)$ , which is minimum when $m=0$ .

This shows that the minimum length of a focal chord is the focal width $\left|4p\right|$ .

Conclusion:

It has been shown that the minimum length of a focal chord is the focal width $\left|4p\right|$ .