Differentiation and proving

Question

Prove that if the curve $y=\frac{x^3}{3} + px + q$ is tangent to the straight line $y = x$, then $4(p-1)^3 + 9q^2 = 0$

I have differentiated the equation of the curve to get $x^2 + p$ and equated it to $1$.

But i have no idea how to proceed to get the subsequent equation.

At what point(s) is the curve tangent to the line? – abiessu Sep 23 '16 at 04:13 — abiessu, Sep 23 '16 at 04:13
Root (1-p) and negative root (1-p)? – Jayne Leblanc Sep 23 '16 at 04:14 — Jayne Leblanc, Sep 23 '16 at 04:14

score 2 · Accepted Answer · answered Sep 23 '16 at 04:33

$$y=\frac{x^3}{3} + px + q$$

Therefore $$y'=x^2 + p $$

Thus the gradient of $y=x$ must be equal to $x^2+p$.

That is $$x^2+p=1$$.

Since $y=x$ at each tangent points $x=\sqrt{1-p}$ and $x=-\sqrt{1-p}$ ,

$$x=\frac{x^3}{3} + px + q$$

$$x^2=\frac{x^4}{3} + px^2 + qx$$

$$(1-p)=\frac{{(1-p)}^2}{3} + p(1-p) + qx$$

$$(1-p)-p(1-p)=\frac{{(1-p)}^2}{3} + qx$$

$$(1-p)(1-p)=\frac{{(1-p)}^2}{3} + qx$$

$$(1-p)^2=\frac{{(1-p)}^2}{3} + qx$$

$$\frac{{2(1-p)}^2}{3} = qx$$

Thus $$\frac{{4(1-p)}^4}{9} = q^2x^2$$

$$\frac{{4(1-p)}^4}{9} = q^2(1-p)$$

For $p \neq 1$ ,

$$4(1-p)^3 = 9q^2$$

$$-4(p-1)^3 = 9q^2$$

$$4(p-1)^3 + 9q^2=0$$

1 Answers1