Review question

# How many common tangents can two parabolas have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7750

## Solution

Let $P$ be the point on the curve $y=ax^2+bx+c$ (where $a$ is non-zero) at which the gradient is $m$. Show that the equation of the tangent at $P$ is $y-mx=c-\frac{(m-b)^2}{4a}.$

The gradient of the quadratic is $\frac{dy}{dx}=2ax+b=m.$

Let the coordinates at $P$ be $P(x_P,y_P)$. Then \begin{align*} x_P &= \frac{m-b}{2a}, \quad \textrm{and} \\ y_P &= a\left(\frac{m-b}{2a}\right)^2+b\left(\frac{m-b}{2a}\right)+c. \end{align*} Thus the tangent at $P$ has the equation $y-\left[a\left(\frac{m-b}{2a}\right)^2+b\left(\frac{m-b}{2a}\right)+c\right] = m\left(x-\frac{m-b}{2a}\right),$ that is, \begin{align*} y-mx &= \frac{(m-b)^2}{4a}+\frac{b(m-b)}{2a}+c-\frac{m(m-b)}{2a}\\ &= c+\frac{1}{4a}(m-b)\left[2b-2m+(m-b)\right]\\ &= c- \frac{(m-b)^2}{4a} \end{align*}

as required.

Show that the curves $y=a_1x^2+b_1x+c_1$ and $y=a_2x^2+b_2x+c_2$ (where $a_1$ and $a_2$ are non-zero) have a common tangent with gradient $m$ if and only if $(a_2-a_1)m^2+2(a_1b_2-a_2b_1)m+4a_1a_2(c_2-c_1)+a_2b_1^2-a_1b_2^2=0.$

The curves have a common tangent with gradient $m$ if and only if the equations of the tangents to the two curves with gradient $m$ are identical. Thus they need the $y$-intercepts to be identical (as they have the same gradients). We have \begin{align} & c_1-\frac{(m-b_1)^2}{4a_1}=c_2-\frac{(m-b_2)^2}{4a_2} \notag\\ \iff & 4a_1a_2c_1-a_2(m-b_1)^2=4a_1a_2c_2-a_1(m-b_2)^2 \notag \\ \iff & (a_2-a_1)m^2+4a_1a_2(c_2-c_1)+2m(a_1b_2-a_2b_1)+a_2b_1^2-a_1b_2^2=0 \quad \label{eq:1} \end{align}

as required.

Show that, in the case $a_1\neq a_2$, the two curves have exactly one common tangent if and only if they touch each other.

If $a_1\neq a_2$ then to get exactly one root of $\eqref{eq:1}$ we need the discriminant of the quadratic to be zero, that is, $4(a_1b_2-a_2b_1)^2=4(a_2-a_1)(4a_1a_2(c_2-c_1)+a_2b_1^2-a_1b_2^2),$ so $-2a_1a_2b_1b_2=4a_1a_2(a_2-a_1)(c_2-c_1)-a_1a_2b_1^2-a_1a_2b_2^2.$ Since $a_1$ and $a_2$ are non-zero by definition, we can cancel the $a_1a_2$ factor: $b_1^2-2b_1b_2+b_2^2=4(a_1-a_2)(c_2-c_1),$ so $(b_2-b_1)^2=4(a_2-a_1)(c_2-c_1).$ The curves touch if and only if there is exactly one solution to $a_1x^2+b_1x+c_1=a_2x^2+b_2x+c_2,$ that is, $(a_2-a_1)x^2+(b_2-b_1)x+(c_2-c_1)=0.$ For one solution we need $(b_2-b_1)^2=4(a_2-a_1)(c_2-c_1)$ which is the same condition for having exactly one common tangent.

In the case $a_1=a_2$, find a necessary and sufficient condition for the two curves to have exactly one common tangent.

If $a_1=a_2$ then there is exactly one solution to $2a_1(b_2-b_1)m+4a_1^2(c_2-c_1)+a_1(b_1^2 - b_2^2)=0.$ This is a linear equation and the only condition needed for there to be a solution for $m$ is $b_1\neq b_2$.

If $b_1 = b_2$, then there are no solutions for $m$ unless $c_1 = c_2$, in which case there are infinitely many solutions.