Finding the Equation for the Line Tangent to a Parabola At a Given Point

The question is as follows:

Verify that the point $A = (8, \frac{25}3)$ lies on the parabola whose focus is (0, 6) and whose directrix is the x-axis. Find an equation for the line that is tangent to the parabola at A.

I was able to verify that the point A lies on the parabola--the simplified equation which I got to be $\frac{x^2}{12} + 3 = y$. However, I am unable to find the equation of the tangent line. I know that the intersection point has to be at A, but I don't know how to find the slope of the line. Any help will be greatly appreciated.

As a side note, I would like it if someone can provide any help without the direct usage of derivatives. Although I am aware that derivatives can be helpful in solving these types of problem, I can't use it because I haven't learned about it in class yet. Thank you.

With recourse to this diagram: it can be seen that the line through the focus and the projection of the given point onto the directrix is perpendicular to the tangent to the given point. Here, these two important points are $(0,6)$ and $(8,0)$ respectively and the line through them has slope $\frac{-6}8=-\frac34$. So the tangent we seek has slope $\frac43$.

Now it remains to solve $\frac{25}3=\frac43\cdot8+c$ for $c$, and it turns out to be $-\frac73$. So the tangent of the parabola at $A$ has equation $3y=4x-7$.

• I am still unsure about how you knew that "the line through the focus and the projection of the given point onto the directrix is perpendicular to the tangent to the given point." Any clarifications on that? – geo_freak Sep 22 '17 at 0:17
• @geo_freak Let focus = F, given point = P and projection of P onto directrix = D. Then the bisector of $\angle FPD$ is the tangent to the parabola - this is the parabola's property. Then consider the isosceles $\triangle FPD$; because this is isosceles (as shown in the diagram), the aforementioned bisector is itself perpendicular to $FD$. My answer hinges on this. – Parcly Taxel Sep 22 '17 at 0:23
• Thank you for the clarification! It helped me understand this so much better. – geo_freak Sep 22 '17 at 1:09

You can use what Fermat called (literally "pre-Calculus") "ab-equation". In order that a line, y= ax+ b, be tangent to a curve, y= f(x), at $x= x_0$, it must "touch" the curve at that line- that is $x_0$ must satisfy f(x)= ax+ b. In fact, for that line to be tangent to that curve, $x_0$ must be a double root. The parabola with focus at (0, 6) and the x-axis as directrix has equation $y= \frac{1}{12}x^2+ 3$. So in order that y= ax+ b be tangent at x= 8 then x= 8 must be a [b]double[/b] root of $\frac{1}{12}x^2+ 3= ax+ b$. That is, we must have $\frac{1}{12}x^2-ax+ 3- b= \frac{1}{12}(x- 8)^2$. Multiplying the right side out, $\frac{1}{12}- ax+ b= \frac{1}{12}x^2- \frac{4}{3}x+ \frac{16}{3}$.

Now, what are a and b?

• It might be worth noting that this depends on the fact that a line intersects a conic in at most two points. For other types of curves the sought-after roots might have other multiplicities. – amd Sep 22 '17 at 0:58

One way is to use the reflective property of the parabola, as in Parcly Taxel’s answer. That’s likely what the writer of the question had in mind.

If you know about pole/polar relationships of lines and points, you can instead use the fact that the polar line of a point on the parabola is the tangent at that point. The referenced article uses a matrix to compute the polar line, but you can also compute it by making the following substitutions into the equation $\frac1{12}x^2-y+3=0$ of the parabola: $x^2\to8x$, $x\to\frac12(x+8)$ and $y\to\frac12\left(y+\frac{25}3\right)$. (If you had a general conic equation, you’d also substitute $y^2\to\frac{25}3y$ and $xy\to\frac12(x+8)\left(y+\frac{25}3\right)$). The resulting equation is $$\frac1{12}(8x)-\frac12\left(y+\frac{25}3\right)+3=\frac23x-\frac12y-\frac{25}6+3=\frac23x-\frac12y-\frac76=0,$$ which can be simplified to $4x-3y=7$ by multiplying it by $6$ to eliminate the denominators.