Question

For each of the equations in Exercises 59–62, y is defined as an implicit function of x. Solve for y, and use what you find to sketch a graph of the equation.

$x^2-3y^2=16$

Short answer

The equation for$y$is$y=\sqrt{\frac{x^2-16}{3}}$

Step-by-step solution

Step 1. Given Information

The given equation is$x^2-3y^2=16$

Step 2. Solve for y

$$\begin{gathered} x^2-3y^2=16 \\ 3y^2=x^2-16 \\ {y}^2=\frac{x^2-16}{3} \\ y=\sqrt{\frac{x^2-16}{3}} \end{gathered}$$

Step 3. Graph of the Equation

We obtained $y=\pm \sqrt{\frac{x^2-16}{3}}$

The graph of hyperbola is shown in the figure