Question

Find the intercepts of the equation $y^2=9+4x^2$.

Short answer

The intercepts of the equation $y^2=9+4x^2$are $(0,3)$and $(0,-3)$.

Step-by-step solution

Step 1. Objective of the Problem

We need to find the intercepts of the equation $y^2=9+4x^2$.

Step 2. Concept

$x-$intercept: The $x-$intercept of the hyperbola is the point on the $x-$axis through which the hyperbola passes.

To find the $x-$intercept of the equation, substitute $y=0$and solve for $x.$

$y-$intercept: The $y-$intercept of the hyperbola is the point on the $y-$axis through which the hyperbola passes.

To find the $y-$intercept of the equation, substitute $x=0$in the equation and solve for$y.$

Step 3. Finding the x-intercept of the equation.

$x-$intercept:

To find the $x-$intercept , substitute $y=0$in the equation $y^2=9+4x^2$.

$$\begin{gathered} y^2=9+4x^2 \\ {0}^2=9+4x^2 \\ 0=9+4x^2 \\ -9=4x^2 \\ {x}^2=\frac{-9}{4} \\ x=\sqrt{\frac{-9}{4}} \end{gathered}$$

Since $x$is not real,the equation $y^2=9+4x^2$has no$x-$intercepts.

Step 4. Finding the y-intercept of the equation.

$y-$intercept: To find the $y-$intercept, substitute $x=0$in the equation $y^2=9+4x^2$

$$\begin{gathered} y^2=9+4x^2 \\ {y}^2=9+4{\left(0\right)}^2{y}^2=9+0 \\ y=\sqrt{9} \\ y=\pm 3 \end{gathered}$$

Therefore, the $y-$intercepts of the equation are$(0,3)$and$(0,-3)$.