Question

refer to the equation $A x^2+C y^2+D x+E y+F=0$, where $A$ and $C$ are not both 0 . Show that if $A C<0$, then the equation is equivalent to an equation of one of the forms $$ \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \quad \text { or } \quad \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1 \text {, } $$ $a>0, b>0$, provided $4 A C F-C D^2-A E^2 \neq 0$.

   refer to the equation $A x^2+C y^2+D x+E y+F=0$, where $A$ and $C$ are not both 0 .
Show that if $A C<0$, then the equation is equivalent to an equation of one of the forms

$$
\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \quad \text { or } \quad \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1 \text {, }
$$

$a>0, b>0$, provided $4 A C F-C D^2-A E^2 \neq 0$.

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Modern Analytic Geometry

Modern Analytic Geometry

William Wooton,… 1st Edition

Chapter 5, Problem 43

Instant Answer

Step 1

We know that \( A \) and \( C \) are not both zero, and we are given the condition \( A C < 0 \). Show more…

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refer to the equation $A x^2+C y^2+D x+E y+F=0$, where $A$ and $C$ are not both 0 . Show that if $A C<0$, then the equation is equivalent to an equation of one of the forms $$ \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \quad \text { or } \quad \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1 \text {, } $$ $a>0, b>0$, provided $4 A C F-C D^2-A E^2 \neq 0$.

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