# Two distinct tangents drawn from an external point to a cubic

The equation $$y=x^3+cx$$ can have two distinct tangents drawn from an external point only if $$c>0$$

I tried to prove this:

Let the external point be (h,k) and let tangents drawn from this point to the cubic $$y=x^3+cx$$ touch at $$(x_1,y_1),(x_2,y_2)$$.$$x_1 \neq x_2$$ or $$y_1 \neq y_2$$. Then $$y_1=x_1^3+cx_1, y_2=x_2^3+cx_2$$

Equation of tangents at those points passing through (h,k):

$$y_1-k=(3x_1^2+c)(x_1-h)$$ and $$y_2-k=(3x_2^2+c)(x_2-h)$$

Solving I got $$h=\frac{2(x_1^2+x_2^2+x_1x_2)}{3(x_1+x_2)}, k=2x_1^2(\frac{x_1^2+x_2^2+x_1x_2}{x_1+x_2})-2x_1^3+\frac{2c(x_1^2+x_2^2+x_1x_2)}{3(x_1+x_2)}$$

Which is correct (I verified with desmos).

For $$c>0$$ these are indeed intersection points of two distinct tangents.

But for $$c<0$$ at first glance from desmos it appears to be intersection point of two distinct tangents again, but on closer observation I noted that these "tangents" were not tangents. They come very close to the cubic but no intersection or tangency point(it becomes sort of like an "asymptotic tangent" at finite points.

Why is this so? No where in my derivation could I find the assumption (directly or indirectly) $$c>0$$.

Also could someone please explain why (h,k) always lies in the region as described in Phillip's answer for the case $$c>0$$?

• Uh... What do we make of the fact that with $c=-1$ there are three tangents from $(1,-0.8)$? (Two also cross the cubic at other, non-tangent points.) – Oscar Lanzi Nov 8 at 16:43
• Oh! So is the original question wrong in the first place? – user600016 Nov 8 at 17:02
• Wrong or missing something. – Oscar Lanzi Nov 8 at 17:23