Question

Show that the graph of an equation of the form

$A{x}^2+C{y}^2+F=0A\ne 0,C\ne 0,F\ne 0$

where A and C are of opposite sign, is a hyperbola with center at (0, 0).

Short answer

The equation of the hyperbola is

$\frac{x^2}{\frac{-F}{A}}+\frac{y^2}{\frac{-F}{C}}=1$

where A and C are of opposite sign, is a hyperbola with center at (0, 0).

Step-by-step solution

Step 1. Given Information

Show that the graph of an equation of the form

$A{x}^2+C{y}^2+F=0A\ne 0,C\ne 0,F\ne 0$

where A and C are of opposite sign, is a hyperbola with center at (0, 0).

Step 2. Hyperbola equation

Equation of hyperbola if the transverse axis is parallel to the x-axis.

$\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1$........(1)

Equation of hyperbola if the transverse axis is parallel to the y-axis.

$\frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1$...... (2)

The center of the hyperbola is in the point (h,k)

Given equation $A{x}^2+C{y}^2+F=0A\ne 0,C\ne 0,F\ne 0$and A and C are of opposite sign

Subtract F from both sides of the equation

role="math" localid="1648264879243" $A{x}^2+C{y}^2=-F$

Divide -F from both sides of the equation

$\frac{A}{-F}{x}^2+\frac{C}{-F}{y}^2=1$

The above equation can be written as

$\frac{x^2}{\frac{-F}{A}}+\frac{y^2}{\frac{-F}{C}}=1$

Now it is given that A and C are of the opposite sign, so the terms $\frac{-F}{A}and\frac{-F}{C}$

are also opposite in sign.

Therefore above equation is an equation of a hyperbola.

Comparing the numerator of the equation of hyperbola with the equation (1) and (2), the center of hyperbola is$(0,0)$therefore$h=0andk=0$.