Question

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation

$x^2-14x+4=9y^2-36y$

Short answer

The standard form of the equation is$\frac{{\left(x-7\right)}^2}{3^2}-\frac{{\left(y-2\right)}^2}{1^2}=1$and represents ahyperbola.

Step-by-step solution

Step 1. Given Information.

Given to write the equation in standard form and to state whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then the equation is to be graphed.

$x^2-14x+4=9y^2-36y$

Step 2. Explanation.

Rewriting the given equation in standard form:

$\begin{array}{l}{x}^2-14x+4=9y^2-36y\\ x^2-14x+49-45=9y^2-36y+36-36\\ {\left(x-7\right)}^2-45=9\left(y^2-4y+4\right)-36\\ {\left(x-7\right)}^2-45=9{\left(y-2\right)}^2-36\\ {\left(x-7\right)}^2-9{\left(y-2\right)}^2=45-36\\ {\left(x-7\right)}^2-9{\left(y-2\right)}^2=9\\ \frac{{\left(x-7\right)}^2}{9}-\frac{{\left(y-2\right)}^2}{1}=1\end{array}$

Hence, the standard form of the equation is role="math" localid="1648402344114" $\frac{{\left(x-7\right)}^2}{3^2}-\frac{{\left(y-2\right)}^2}{1^2}=1$

This equation is of the form role="math" localid="1648402366060" $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$which represents a hyperbola.

Hence, the graph of the equation is a hyperbola.

Graphing the equation:

Step 3. Conclusion.

Hence, the standard form of the equation is$\frac{{\left(x-7\right)}^2}{3^2}-\frac{{\left(y-2\right)}^2}{1^2}=1$and represents a hyperbola.