A parabola with a line and a circle; the tangent to a parabola

The points common to a line and a parabola are the solutions of the system of their equations. Put the line into the parabola and get a quadratic equation: two solutions – the line cuts; one double solution – the line is a tangent; no solution – it misses.

An example: $\displaystyle {y^{2}}={4x}$ and $\displaystyle {y}={2x-4}$.

$\displaystyle {x}={1}$ or $\displaystyle {x}={4}$, and from the line: $\displaystyle (1,-2)$ and $\displaystyle (4,4)$.

Careful: a line parallel to the axis of symmetry, such as $\displaystyle {y}={3}$, meets the parabola at exactly one point – $\displaystyle \left(\frac{9}{4},3\right)$ – but it crosses it and is no tangent. Here the equation after substituting is of the first degree, not a quadratic with a double solution.

The tangent to the parabola $\displaystyle {y^{2}}={2px}$ at the point $\displaystyle (x_0,y_0)$ on it:

Where the formula comes from: the upper half of the parabola is the function $\displaystyle {y}={\sqrt{2px}}$, and its derivative at the point is $\displaystyle {y'}={\frac{2p}{2\sqrt{2px}}}={\frac{p}{y_0}}$. The tangent: $\displaystyle {y-y_0}={\frac{p}{y_0}(x-x_0)}$. Multiply by $\displaystyle y_0$: $\displaystyle {y\cdot y_0-y_0^{2}}={px-px_0}$, and put $\displaystyle 2px_0$ in place of $\displaystyle y_0^{2}$ – the point is on the parabola. This gives $\displaystyle {y\cdot y_0}={p(x+x_0)}$. On the lower half, $\displaystyle {y}={-\sqrt{2px}}$, the working is the same.

An example: $\displaystyle {y^{2}}={8x}$, that is $\displaystyle {p}={4}$, at the point $\displaystyle (2,4)$: $\displaystyle {4y}={4(x+2)}$, that is $\displaystyle {y}={x+2}$.

A circle and a parabola, too, meet at the solutions of the system. It is convenient to put $\displaystyle y^{2}$ from the parabola into the circle’s equation.

An example: $\displaystyle {x^{2}+y^{2}}={12}$ and $\displaystyle {y^{2}}={4x}$.

• $\displaystyle {x^{2}+4x}={12}$, so $\displaystyle {x}={2}$ or $\displaystyle {x}={-6}$.
• $\displaystyle {x}={-6}$ is rejected: $\displaystyle {y^{2}}={4\cdot(-6)<0}$. On this parabola $\displaystyle x\ge0$.
• $\displaystyle {x}={2}$: $\displaystyle {y^{2}}={8}$, so $\displaystyle {y}={2\sqrt{2}}$ or $\displaystyle {y}={-2\sqrt{2}}$.

Two common points: $\displaystyle (2,2\sqrt{2})$ and $\displaystyle (2,-2\sqrt{2})$ – symmetric about the $\displaystyle x$-axis, like both shapes.

Find the points common to the parabola $\displaystyle {y^{2}}={6x}$ and the line $\displaystyle {2x-y}={6}$.

Mark the points on the board.

Write the tangents to the parabola $\displaystyle {y^{2}}={12x}$ at the points $\displaystyle (3,6)$ and $\displaystyle (3,-6)$.

Where do the two tangents meet? What is special about that point?

For which value of $\displaystyle n$ does the line $\displaystyle {y}={x+n}$ touch the parabola $\displaystyle {y^{2}}={8x}$?

Find the points common to the circle $\displaystyle {x^{2}+y^{2}}={5}$ and the parabola $\displaystyle {y^{2}}={4x}$.

The line $\displaystyle {y}={-4}$ meets the parabola $\displaystyle {y^{2}}={8x}$ at one point. Is it a tangent to it? Explain.