# Finding an equation of two perpendicular tangent lines of a parabola

A parabola would be given as the following: $y^2=4px$.

1) The question is, one wishes to find each equation for two orthogonal (perpendicular) tangent lines of a parabola.

What would be the equations?

Add: And one wishes to find the locus of the intersecting point of two orthogonal tangent lines. How would one be able to get the locus?

The book I am reading to says that $y=mx+\frac{p}{m}$ can be the equation for a tangent line of a parabola. Is this right?

2) And suppose that there is a point $P(x_0,y_0)$ outside the parabola (this parabola is the aforementioned.). Assume that from the point, two tangent lines can be drawn. For each line, then, there would be points $Q_1(x_1,y_1)$, $Q_2(x_2,y_2)$.

Then why is $y_1y=2p(x+x_1)$ and $y_2y=2p(x+x_2)$?

Edit: I'll add one more question to 1):

• For the equations in your second question, what are "$x$" and "$y$", and where are "$x_0$" and "$y_0$"? – Blue Oct 14 '12 at 15:55

I'm assuming you don't want any calculus involved here (too bad!), so let $\,(a,b)\,,\,(c,d)\,$ be the points on the parabola through which pass two tangent lines to it that are perpendicular:

$$\text{First tangent: we need to solve the system}\;\;\;\;y^2=4px\;\;,\;\;y-b=m(x-a)\Longrightarrow$$

$$(m(x-a)+b)^2=4px$$

$$\text{Second tangent: we need to solve the sytem}\;\;\;\;y^2=4px\;\;,\;\;y-d=-\frac{1}{m}(x-c)\Longrightarrow$$

$$\left(-\frac{1}{m}(x-c)+d\right)^2=4px$$

Of course, solving the above take into account that

$$b^2=4pa\;\;,\;\;d^2=4pc$$

since both points were chosen to be on the parabola.

Also, remember that two straight lines (none of which is horizontal/vertical) with slopes $\,m_1\,,\,m_2\,$ are perpendicular iff $\,m_1m_2=-1\,$, and this the reason we took the second tangent's slope to be $\,-1/m\,$

• How would usage of calculus simplify the matter? – rrqq Oct 14 '12 at 13:16
• $$y^2=4px\Longrightarrow 2yy'=4p\Longrightarrow y'=\frac{2p}{y}$$ and $\,y'\,$ is the slope of the tangent line in the chosen point. If you do the above with two generic points of the parabola and then use that it must be that ths slope of one must be minus the inverse of the other slope then you can make it easier, I guess. – DonAntonio Oct 14 '12 at 13:21

The way to answer to the questions is to ensure that the lines are tangent to the parabola, i.e. they have the same slope as the parabola in the points of contact (this is the missing piece of the answers so far).

To make things easier let's swap the coordinates: $$x = {y^2 \over 4p}$$

The slope of the parabola is it derivative: $$x' = {2y \over 4p} = {y \over 2p}$$

Now let's consider the tangents. We can have two generic equations with the following parameters: $$slope = m$$ and $$coefficient = q_1$$ for the first equation, $$slope = {-1 \over m}$$ (perpendicular to m) and $$coefficient = q_2$$ for the second equation.

The equations will be: $$x = {my + q_1}$$ $$x = {-\frac{1}{m}y + q_2}$$

At this point we equate the slope of the parabola to that of the first equation: $$y = 2mp$$ and substituting that in the equation of the parabola: $$x = m^2p.$$

Repeating the process for the second equation: $$y = {-2p \over m}$$ and substituting that in the equation of the parabola again: $$x = {p \over m^2p}.$$

Substituting those values in the respective equations of the lines ensures that we find the coefficients of the tangents. For the first equation: $$m^2p = {2m^2p + q_1}$$ $$q_1 = -m^2p$$ and for the second equation: $$q_2 = {-p \over m^2}.$$

In total the equations of the tangents are: $$x = my - m^2p$$ $$x = {-1\over m} y - {p \over m^2}.$$

[Edit: Greatly simplified over previous version. Plus, corrected a mis-reading of the equation.]

Parabola Fun Fact:

If a parabola with a horizontal axis has a chord with endpoints $A(x_1, y_1)$ and $B(x_2, y_2)$, then the tangent line through the point with $y$-coordinate $\frac{1}{2}(y_1+y_2)$ is parallel to $AB$.

Now, the slope of $AB$ is

$$m = \frac{y_1-y_2}{x_1-x_2} = \frac{y_1-y_2}{\frac{1}{4p}y_1^2-\frac{1}{4p}y_2^2}=\frac{4p}{y_1+y_2}$$

By the Parabola Fun Fact, $AB$ is parallel to the line tangent to the parabola at point $T$, where

$$T = \left( \frac{1}{4p} \left(\frac{y_1+y_2}{2}\right)^2, \frac{y_1+y_2}{2}\right) = \left( \frac{\left(y_1+y_2\right)^2}{16p}, \frac{y_1+y_2}{2}\right) = \left( \frac{p}{m^2}, \frac{2p}{m}\right)$$

We determine the equation of the tangent line using the point-slope form, $y-y_0=m(x-x_0)$: \begin{align} y - \frac{2p}{m} = m \left(x-\frac{p}{m^2}\right) \qquad \to \qquad y = mx+\frac{p}{m} \end{align}

For the perpendicular tangent, simply replace $m$ by $-1/m$: $$y = -\frac{1}{m}x-pm$$

For the locus of points of intersection, simply solve the system of the two tangent-line equations. The result is very nice.