Tangent of two parabolas

Question

Please, help. We have two parabolas:

$$1.\;\; y^2 = x$$ $$2. \;\;y = 2x^2$$ How to calculate common tangent line for them?

Thank you for your help.

Answer 1

Hint...Let the tangent be $y=mx+c$.

Solve this simultaneously with the first parabola to create a quadratic which must have double roots. Apply the condition that the discriminant is zero to make an equation for $m$ and $c$.

Repeat this with the second parabola and solve the simultaneous equations for $m$ and $c$.

Answer 2

Let $A(a^2,a)$ and $B(b,2b^2)$ be the points on $C_1:y^2=x$ and $C_2:y=2x^2$ respectively.

Equation of tangent of $C_1$ at $A$:

\begin{align*} ay &= \frac{x+a^2}{2} \\ y &= \frac{x}{2a}+\frac{a}{2} \quad \cdots \cdots \: (1) \end{align*}

Equation of tangent of $C_2$ at $B$:

\begin{align*} \frac{y+2b^2}{2} &= 2bx \\ y &= 4bx-2b^2 \quad \cdots \cdots \: (2) \end{align*}

Comparing $(1)$ and $(2)$,

$$ \left \{ \begin{array}{rcl} \frac{1}{2a} &=& 4b \\ \frac{a}{2} &=& -2b^2 \\ \end{array} \right.$$

$$\frac{1}{4}=-8b^3 \implies b=-\frac{1}{\sqrt[3]{32}}$$

The common tangent is

$$y=-\sqrt[3]{2} x-\frac{1}{\sqrt[3]{128}}$$

Useful fact:

Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x',y')$ is given by

$$ax'x+h(y'x+x'y)+by'y+g(x+x')+f(y+y')+c=0$$

Answer 3

It's easy to see that the tangent to the second parabola at $(t,2t^2)$ is $$ y-2t^2=4t(x-t) $$ that is, $$ y=4tx-2t^2 $$ Substituting into the first one, we get $$ (4tx-2t^2)^2=x $$ that becomes $$ 16t^2x^2-(16t^3+1)x+4t^4=0 $$ which must have zero discriminant: $$ (16t^3+1)^2-16^2t^6=0 $$ Solve and substitute.