Chapter 8.1, Problem 76E

Solution

(a.)

To Prove: For a parabola, the two end points of the latus rectum and the point of intersection of the axis and directrix are the vertices of an isosceles right triangle.

It has been shown that for a parabola, the two end points of the latus rectum and the point of intersection of the axis and directrix are the vertices of an isosceles right triangle.

Given:

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

Concept used:

The focal chord of a parabola perpendicular to the axis of the parabola is the latus rectum.

Calculation:

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

The axis of this parabola is the $y$ axis.

The directrix of this parabola is $y=-p$ .

Then, the point of intersection of the axis and directrix of the parabola, is the point $\left(0,-p\right)$ .

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.

Now, the slope of the latus rectum, which is perpendicular to the axis of the parabola and hence perpendicular to the $y$ axis, is $m=0$ .

Put $m=0$ in $x=2p\left(m\pm \sqrt{m^2+1}\right)$ to get,

  $x=\pm 2p$

Put $x=\pm 2p$ in the equation of the parabola; $x^2=4py$ to get,

  $y=p$

Hence, the two end points of the latus rectum are $\left(2p,p\right)$ and $\left(-2p,p\right)$ .

Now, clearly the points $A:\left(0,-p\right)$ , $B:\left(2p,p\right)$ and $C:\left(-2p,p\right)$ are the vertices of a triangle.

Applying the distance formula,

  $\begin{array}{c}AB=\sqrt{{\left(2p-0\right)}^2+{\left(p-\left(-p\right)\right)}^2}\\ =\sqrt{{\left(2p\right)}^2+{\left(2p\right)}^2}\\ =\sqrt{2{\left(2p\right)}^2}\\ =2\sqrt{2}p\end{array}$

Similarly,

  $\begin{array}{c}AC=\sqrt{{\left(-2p-0\right)}^2+{\left(p-\left(-p\right)\right)}^2}\\ =\sqrt{{\left(-2p\right)}^2+{\left(2p\right)}^2}\\ =\sqrt{2{\left(2p\right)}^2}\\ =2\sqrt{2}p\end{array}$

This implies that $AB=AC$ .

Applying the two-point slope formula, the slope of $AB$ is,

  $\begin{array}{c}m=\frac{p-\left(-p\right)}{2p-0}\\ =\frac{p+p}{2p}\\ =\frac{2p}{2p}\\ =1\end{array}$

Similarly, the slope of $AC$ is,

  $\begin{array}{c}m=\frac{p-\left(-p\right)}{-2p-0}\\ =\frac{p+p}{-2p}\\ =\frac{2p}{-2p}\\ =-1\end{array}$

So, the product of the slope of $AB$ and the slope of $AC$ is $-1$ .

This shows that $AB$ is perpendicular to $AC$ .

Now, $A:\left(0,-p\right)$ , $B:\left(2p,p\right)$ and $C:\left(-2p,p\right)$ are the vertices of a triangle such that $AB=AC$ and $AB$ is perpendicular to $AC$ .

Thus, $A:\left(0,-p\right)$ , $B:\left(2p,p\right)$ and $C:\left(-2p,p\right)$ are the vertices of an isosceles right triangle.

This shows that for a parabola, the two end points of the latus rectum and the point of intersection of the axis and directrix are the vertices of an isosceles right triangle.

Conclusion:

It has been shown that for a parabola, the two end points of the latus rectum and the point of intersection of the axis and directrix are the vertices of an isosceles right triangle.

(b.)

To Prove: The legs of the isosceles right triangle obtained in part (a) are tangent to the parabola.

It has been shown that the legs of the isosceles right triangle obtained in part (a) are tangent to the parabola.

Given:

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

Concept used:

The focal chord of a parabola perpendicular to the axis of the parabola is the latus rectum.

Calculation:

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

The axis of this parabola is the $y$ axis.

The directrix of this parabola is $y=-p$ .

Then, the point of intersection of the axis and directrix of the parabola, is the point $\left(0,-p\right)$ .

As determined previously, the two end points of the latus rectum are $\left(2p,p\right)$ and $\left(-2p,p\right)$ .

As shown in part (a), $A:\left(0,-p\right)$ , $B:\left(2p,p\right)$ and $C:\left(-2p,p\right)$ are the vertices of an isosceles right triangle such that $AB=AC$ and $AB$ is perpendicular to $AC$ .

As shown previously, a line tangent to the given parabola at the point $\left(a,b\right)$ crosses the $y$ axis at $\left(0,-b\right)$ .

This implies that the tangent to the given parabola at the point $B:\left(2p,p\right)$ crosses the $y$ axis at $\left(0,-p\right)$ , which is precisely the point $A$ .

Similarly, the tangent to the given parabola at the point $C:\left(-2p,p\right)$ crosses the $y$ axis at $\left(0,-p\right)$ , which is again precisely the point $A$ .

This implies that $AB$ and $AC$ are tangent to the given parabola.

This shows that the legs of the isosceles right triangle obtained in part (a) are tangent to the parabola.

Conclusion:

It has been shown that the legs of the isosceles right triangle obtained in part (a) are tangent to the parabola.