A high-school question on Analytic Geometry.

Question

Find and identify the locus of the incident tangents to a parabola $y^2=2px$ that their angle between them is $45°$.

I tried to use the formula for tangents, i.e if $m_1, m_2$ are the slopes of these tangents then we have the equation: $$\tan (45°)=\frac{m_1-m_2}{1+m_1m_2} $$ where $m_i= \pm\frac{\sqrt{2p}}{2\sqrt{x}}$.

The problem is that the answer in the book, is that the equation of the locus is $(x+1.5p)^2-y^2=2p^2$. And I don't get this answer I get from the equation above that $x^2-3px+p^2/4=0$, which doesn't resemble the answer in the book.

Can someone help me with this question?

Thanks!

Answer 1

Let $(x_1, y_1)$ be a point of tangency on the parabola $y^2 = 2 p x $, then

$ x_1 = \dfrac{ y_1^{2}}{2p} $

So the point of tangency is $\left( \dfrac{y_1^2}{2p}, y_1 \right) $. By implicit differentation, $ 2 y \dfrac{dy}{dx} = 2 p $, so the slope at the point of tangency is

$$m = \dfrac{dy}{dx} = \dfrac{p}{y_1} $$

Since the tangent passes through the point on the locus $(x, y)$ then

$ y - y_1 = \dfrac{p}{y_1} \bigg(x - \dfrac{y_1^2}{2p}\bigg) $

Cross multiply,

$ y_1 (y - y_1) = p x - \dfrac{1}{2} y_1^2 $

Re-arrange,

$ \dfrac{1}{2} y_1^2 - y_1 y + p x = 0 $

So the solutions are

$ y_1 = y \pm \sqrt{ y^2 - 2 p x } $

Let

$ y_1^+ = y + \sqrt{y^2 - 2 p x} $

$y_1^- = y - \sqrt{y^2 - 2 p x} $

And let the slopes at these two points be $m_1 = \tan(\theta_1) = \dfrac{p}{y_1^+} $ and $m_2 = \tan(\theta_2) = \dfrac{p}{y_1^-} $, then

$45^\circ = \theta_1 - \theta_2 $

So that,

$ \tan 45^\circ = 1 = \tan(\theta_1 - \theta_2) = \dfrac{m_1 - m_2}{1 +m_1m_2} $

From which

$ 1 + m_1 m_2 = m_1 - m_2 $

Hence,

$ 1 + \dfrac{p^2}{y_1^+ y_1^-} = p \bigg( \dfrac{1}{y_1^+} - \dfrac{1}{y_1^-} \bigg)$

So

$ y_1^+ y_1^- + p^2 = p (y_1^- - y_1^+)$

And this gives us

$ y^2 - (y^2 - 2 p x) + p^2 = p ( - 2 \sqrt{ y^2 - 2 p x } )$

This simplifies to

$ 2 x + p = - 2 \sqrt{y^2 - 2 p x} $

(This means that we must have $ x \lt -\dfrac{p}{2} $)

Square both sides,

$ 4 x^2 + 4 x p + p^2 = 4 (y^2 - 2 p x) $

Re-arrange,

$ 4 x^2 + 12 x p + p^2 - 4 y^2 = 0 $

Divide through by $4$,

$ x^2 + 3 x p + \dfrac{p^2}{4} - y^2 = 0 $

Complete the square in $x$,

$ \bigg( x + \dfrac{ 3 p }{2} \bigg)^2 - \dfrac{9 p^2}{4} + \dfrac{p^2}{4} - y^2 = 0 $

which finally gives us,

$ \bigg( x + \dfrac{ 3 p }{2} \bigg)^2 - y^2 = 2 p^2 $

Answer 2

Let $(x, y)=(2pt^2, 2pt)$,

then $\frac{dy}{dx}=\frac{1}{2t}$

The equation of the tangent at $(2pt^2, 2pt)$ is:

$$2pt^2-2ty+x=0 \tag{1} $$

Thus if $D(x, y)$ is a point on the locus and $[t_1], [t_2]$ are the points of contact, $t_1, t_2$ are the roots of (1).

Hence $$t_1+t_2=\frac{y}{p} \tag{2}$$

$$t_1t_2=\frac{x}{2p} \tag{3}$$

Furthermore, since the tangents cut at $45^o$,

$$\frac{|\frac{1}{2t_1}-\frac{1}{2t_2}|}{|1+\frac{1}{4t_1t_2}|}=1$$

$$\frac{2|t_1-t_2|}{|1+4t_1t_2|}=1$$

$$4\left(t_1-t_2 \right)^2 =\left(1+4t_1t_2 \right)^2 $$

$$4\left[(t_1+t_2)^2-4t_1t_2 \right] =\left(1+4t_1t_2 \right)^2 \tag{4} $$ Eliminating $t_1$ and $t_2$ in (4) by using (2) and (3), and arranging terms give the desired result.