An ellipse with a line and a circle

As with a circle: put the line into the ellipse’s equation, and the number of solutions of the quadratic is the number of common points – two, one (the line is a tangent) or none.

An example: $\displaystyle {\frac{x^{2}}{25}+\frac{y^{2}}{9}}={1}$ and $\displaystyle {y}={x-3}$.

• Multiply by $\displaystyle 225$: $\displaystyle {9x^{2}+25y^{2}}={225}$, and substitute: $\displaystyle {9x^{2}+25(x-3)^{2}}={225}$.
• $\displaystyle {34x^{2}-150x}={0}$, that is $\displaystyle {x(34x-150)}={0}$.
• $\displaystyle {x}={0}$, $\displaystyle {y}={-3}$ – the vertex $\displaystyle (0,-3)$; or $\displaystyle {x}={\frac{75}{17}}$, $\displaystyle {y}={\frac{24}{17}}$.

A line touches an ellipse when the quadratic has one solution – its discriminant is $\displaystyle 0$.

An example: for which values of $\displaystyle n$ does the line $\displaystyle {y}={x+n}$ touch the ellipse $\displaystyle {\frac{x^{2}}{4}+y^{2}}={1}$?

• $\displaystyle {x^{2}+4(x+n)^{2}}={4}$, that is $\displaystyle {5x^{2}+8nx+4n^{2}-4}={0}$.
• The discriminant: $\displaystyle {64n^{2}-20(4n^{2}-4)}={80-16n^{2}}$.
• $\displaystyle {80-16n^{2}}={0}$, so $\displaystyle {n}={\sqrt{5}}$ or $\displaystyle {n}={-\sqrt{5}}$: two parallel tangents, one on each side.

A circle with its centre on the $\displaystyle x$-axis and a canonical ellipse: in both equations $\displaystyle y$ appears only as $\displaystyle y^{2}$. Isolate $\displaystyle y^{2}$ in one and put it into the other, and get an equation in $\displaystyle x$ alone.

An example: $\displaystyle {2x^{2}+4y^{2}}={16}$ and $\displaystyle {(x-2)^{2}+y^{2}}={4}$.

• From the circle: $\displaystyle {y^{2}}={4-(x-2)^{2}}={4x-x^{2}}$.
• In the ellipse: $\displaystyle {2x^{2}+4(4x-x^{2})}={16}$, that is $\displaystyle {x^{2}-8x+8}={0}$, and $\displaystyle {x}={4\pm2\sqrt{2}}$.
• $\displaystyle {x}={4+2\sqrt{2}\approx6.8}$ is rejected: on the ellipse $\displaystyle x^{2}\le8$. Indeed, $\displaystyle {y^{2}}={4x-x^{2}}$ comes out negative.
• $\displaystyle {x}={4-2\sqrt{2}\approx1.17}$: $\displaystyle {y^{2}}={4x-x^{2}>0}$, so two points, symmetric about the $\displaystyle x$-axis.

Careful: a solution of the equation in $\displaystyle x$ gives a point only if the $\displaystyle y^{2}$ it gives is not negative.

Find the points common to the ellipse $\displaystyle {\frac{x^{2}}{16}+\frac{y^{2}}{4}}={1}$ and the line $\displaystyle {x+2y}={4}$.

The ellipse $\displaystyle {\frac{x^{2}}{8}+\frac{y^{2}}{2}}={1}$ and the line $\displaystyle {y}={-\frac{1}{2}x+2}$.

How do the line and the ellipse lie?

Mark on the board the points common to the ellipse and the line.

For which values of $\displaystyle n$ does the line $\displaystyle {y}={2x+n}$ touch the ellipse $\displaystyle {x^{2}+\frac{y^{2}}{4}}={1}$?

Find the points common to the ellipse $\displaystyle {\frac{x^{2}}{25}+\frac{y^{2}}{9}}={1}$ and the circle $\displaystyle {x^{2}+y^{2}}={16}$.

The ellipse $\displaystyle {2x^{2}+4y^{2}}={16}$ and the circle $\displaystyle {(x-2)^{2}+y^{2}}={m^{2}}$, with $\displaystyle m>0$.

The distances of $\displaystyle (2,0)$ from the ellipse’s vertices $\displaystyle (\pm2\sqrt{2},0)$ are $\displaystyle 2\sqrt{2}-2$ and $\displaystyle 2\sqrt{2}+2$. For which values of $\displaystyle m$ do the two curves have exactly one common point? And for which – two points, or none? Use the board.

Write circles of your own on the board, $\displaystyle {(x-2)^{2}+y^{2}}={m^{2}}$ with different $\displaystyle m$, and see how the number of common points changes.