How to find the tangent lines of a parabola which must pass through a point without calculus.

Question

I‘m tutoring a student who doesn’t know calculus (yet). I was given the following question:

Find all lines of tangency of the graph $y=x^2$ which pass through the point $P(-6,-5)$.

I know I have to use $\sqrt{b^2-4ac}=0$, but I end up with too many unknown variables. Any suggestions?

I've answered a very similar question question before: https://math.stackexchange.com/a/2833210/ . — Xander Henderson, Sep 08 '23 at 19:30
Joachimsthal notations give $((-5)-(-6)^2)(y-x^2)-((y+(-5))/2-(-6)x)^2=\5\cdot((x+6)-\frac{6+\sqrt{41}}{10}(y+5))\cdot((x+6)-\frac{6-\sqrt{41}}{10}(y+5))=\ (-\frac14)\cdot((y+5)-(2 (-6+\sqrt{41}))(x+6))\cdot\((y+5)-(2 (-6-\sqrt{41}))(x+6))$ — Jan-Magnus Økland, Sep 08 '23 at 20:56

score 3 · Answer 1 · answered Sep 08 '23 at 19:27

3

The equation of the line throught $P$ is: $$y=mx+6m-5$$ Equating with$y=x^2$: $$x^2-mx-6m+5=0\implies x=\frac{m\pm \sqrt{m^2-4\cdot(5-6m)}}{2}$$ To be tangent, discriminant has to be $0$: $$m^2+24m-20=0$$ Can you finish it now?

answered Sep 08 '23 at 19:27

Matteo

6,661

Yes, thank you! – Mimir2902 Sep 09 '23 at 20:12

score 3 · Accepted Answer · answered Sep 08 '23 at 19:29

3

Let the line be $y+6 = m(x+5)$.

Then, as you are implying, the graph of $y=x^2 - (m(x+6) - 5)$ should be tangent to the $x$-axis, so its vertex should happen at $y=0$.

This equation rewrites as $y = x^2 - mx + (5-6m)$.

Putting this parabola in vertex form gives us $\displaystyle y = x^2 - mx + \frac{m^2}{4}+(5-6m-\frac{m^2}{4})$

From this, we see that the solutions for $m$ are those where $\displaystyle 5-6m-\frac{m^2}{4}=0$

answered Sep 08 '23 at 19:29

Saketh Malyala

13,897

Makes sense, thank you! – Mimir2902 Sep 09 '23 at 20:11

How to find the tangent lines of a parabola which must pass through a point without calculus.

2 Answers2