Conic sectionsAQA A-Level Further Maths: Revision notes

The curve $y^2=4ax$ with $a>0$ is a parabola opening to the right. Its vertex is at the origin, the $x$-axis is its line of symmetry, and it exists only for $x\ge0$ (since $y^2\ge0$). It meets the axes only at the origin. The point $(a,2a)$ and its reflection $(a,-2a)$ lie on it. For $y^2=12x$, $4a=12$ so $a=3$, and the curve passes through $(3,6)$ and $(3,-6)$. For $a<0$ the parabola opens to the left.

The ellipse is a closed oval centred at the origin, symmetrical in both axes.

• $x$-intercepts: put $y=0$, giving $x=\pm a$, so $(\pm a,0)$.
• $y$-intercepts: put $x=0$, giving $y=\pm b$, so $(0,\pm b)$.
• If $a>b$ the longer axis is horizontal; if $a=b$ the ellipse is a circle of radius $a$. Example: $\frac{x^2}{25}+\frac{y^2}{9}=1$ meets the axes at $(\pm5,0)$ and $(0,\pm3)$. The point $\left(4,\frac95\right)$ lies on it because $\frac{16}{25}+\frac{9}{25}=1$.

The hyperbola has two separate branches, one on each side of the $y$-axis, and is symmetrical in both axes.

• $x$-intercepts: $(\pm a,0)$.
• No $y$-intercepts: $x=0$ would give $y^2=-b^2$, which has no real solution.
• Asymptotes: for large $|x|$ and $|y|$ the equation approaches $\frac{x^2}{a^2}=\frac{y^2}{b^2}$, so the asymptotes are $y=\pm\frac bax$. Example: $\frac{x^2}{16}-\frac{y^2}{9}=1$ has intercepts $(\pm4,0)$ and asymptotes $y=\pm\frac34x$. The branches approach, but never cross, these lines.

For $xy=c^2$ the asymptotes are the coordinate axes, $x=0$ and $y=0$. The curve has two branches, in the first and third quadrants, and no intercepts with either axis. It is symmetrical in the lines $y=x$ and $y=-x$. Example: $xy=9$ passes through $(3,3)$, $(1,9)$, $(9,1)$, $(-3,-3)$. It is called rectangular because its asymptotes are at right angles.

When sketching any of these curves, mark: the intercepts with the axes (put $x=0$, then $y=0$), the asymptotes as dashed lines, the symmetry, and the overall shape. Label each intercept with coordinates. To find where a conic meets a line, substitute the line into the equation and solve the resulting quadratic. Its discriminant decides whether there are two, one or no intersections. For example, $x+y=k$ meets $xy=9$ when $x^2-kx+9=0$, so there are two points when $k^2>36$ and exactly one when $k=\pm6$.