Question

Reduce the conic x2 − 6xy+y2 − 4 = 0 to standard form. Hence the given conic.

Answer 1

Answer on Question #64857 – Math – Linear Algebra

Question

Reduce the conic $x^{2} - 6xy + y^{2} - 4 = 0$ to standard form. Hence the given conic.

Solution

$$x^{2} - 6xy + y^{2} - 4 = 0 \Rightarrow (x^{2} - 6xy + 9y^{2}) - 9y^{2} + y^{2} - 4 = 0 \Rightarrow (x - 3y)^{2} - 8y^{2} - 4 = 0.$$

Now we have

$$(x - 3y)^{2} - 8y^{2} = 4.$$

Substituting $X = x - 3y$ and $Y = y$ into (1) we will obtain the following equation:

$$X^{2} - 8Y^{2} = 4$$

Dividing by 4

$$\frac{X^{2}}{4} - \frac{Y^{2}}{\frac{1}{2}} = 1.$$

This is a canonical equation of hyperbola.

Answer: $\frac{X^2}{4} - \frac{Y^2}{\frac{1}{2}} = 1$; the given conic is hyperbola.

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