Recognising a conic

Statement

Recognise the conic $x^2 - 4y^2 - 2x - 8y - 7 = 0.$

Solution

The coefficients of the quadratic terms are $A=1$ and $B=-4$: they have opposite signs, so the conic is a hyperbola. We identify the centre and its elements by completing the square.

We group and complete: $x^2-2x = (x-1)^2-1,$ $-4y^2-8y = -4(y^2+2y) = -4\bigl[(y+1)^2-1\bigr] = -4(y+1)^2+4.$ Substituting: $(x-1)^2-1-4(y+1)^2+4-7=0 \;\Longrightarrow\; (x-1)^2-4(y+1)^2=4.$ Dividing by $4$: $\boxed{\frac{(x-1)^2}{4}-\frac{(y+1)^2}{1}=1.}$ It is a hyperbola with centre $(1;-1)$, with $a^2=4$ ($a=2$) and $b^2=1$ ($b=1$). Since $a\ne b$ it is not rectangular; its asymptotes are $y+1=\pm\frac{1}{2}(x-1).$

Links

Topics: Homographic hyperbola
Concepts: Asymptote · Completing the square · Conic · Hyperbola
Functions: Hyperbola
Skills: Analytic geometry · Reasoning by cases
Exercise type: Geometric problem