Can we find the area of the triangle created by a common tangent?

Can we find the area of the triangle created by a common tangent?

Question

Prove that the curves $y^2 = 16x$ and $y = x^2 + 3$ touch at the point $(1,4)$. Prove also that the common tangent at this point forms with the axes a triangle of unit area.

Suggestion

The important word here is touch.

Suppose two curves both go though a point $P$. What can we say about the tangent to each curve at $P$ if the curves touch at $P$?

Solution

To show that the two curves touch one another at $(1,4)$, we’ll show that

1. this point lies on both curves, and that

---

1. their gradients are equal at this point.

For the curve $y = x^2 + 3$, since $4 = 1^2 + 3$, the point $(1,4)$ lies on the curve.

We also have that $y' = 2x \implies y'(1) = 2$.

For the curve defined by $y^2 = 16x$, we have that $4^2 = 16 \times 1$, and so the point $(1, 4)$ lies on this curve too.

If

$y^2 = 16x$

, then

$y = 4\sqrt{x}=4x^{\frac{1}{2}}$

, so

$$\begin{align*} \frac{dy}{dx} &= 4\times\frac{1}{2}\times x^{-\frac{1}{2}} \\ &=\frac{2}{\sqrt{x}} \end{align*}$$

which gives that $y'(1)=2$.

Thus at $(1,4), y' = 2$ for both curves, and they do indeed kiss at $(1,4)$.

Alternatively, the curves intersect when $(x^2+3)^2=16x$, or $x^4+6x^2-16x+9=0$.

This factorises to $(x-1)^2(x^2+2x+9)=0$ (we are looking for a double factor of $(x-1)$, since the question tells us to expect a touching at $x = 1$).

Thus there is a double root at $x = 1$, and so the curves touch at $(1,4)$.

We might note the extra information that $x^2+2x+9=0$ has no roots, since its discriminant is negative, so $(1,4)$ is the only place these curves coincide.

The tangent of the two curves at $(1,4)$ is of the form $y = mx + c$, where $m$ is the gradient of both curves at this point, and $c$ is such that $(1,4)$ lies on the line.

We know that $m = 2$. As $c$ satisfies $4 = m \times 1 + c$, we have $c = 4 - 2 \times 1 = 2$. So the tangent is $y = 2x + 2.$

This intersects the axes at $(-1,0)$ and $(0,2)$, and so the area of the required triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1.$