Question

The intercepts of the equation$x^2+4y^2=16$ are ______________________________.

Short answer

The intercepts are $\left(4,0\right),\left(-4,0\right),\left(0,2\right)\mathrm{and}\left(0,-2\right)$.

Step-by-step solution

Step 1. Given information

The given equation is $x^2+4y^2=16$.

Need to find the intercepts of the given equation.

Step 2. Finding y-intercept

Substitute $x=0$in the equation $x^2+4y^2=16$ and solve for yto obtain the y-intercepts.

$\begin{array}{rcl}{\left(0\right)}^2+4{\left(y\right)}^2& =& 16\\ 4y^2& =& 16\\ y^2& =& 4\\ y^2-4& =& 0\\ \left(y-2\right)\left(y+2\right)& =& 0\end{array}$

$y=2ory=-2$

The intercepts on the y-axis are $\left(0,2\right)$and $\left(0,-2\right)$.

Step 3. Finding x-intercept

Substitute $y=0$in the equation $x^2+4y^2=16$and solve for$x$to obtain the x-intercepts.

$\begin{array}{rcl}{x}^2+4{\left(0\right)}^2& =& 16\\ x^2& =& 16\\ x^2-16& =& 0\\ \left(x-4\right)\left(x+4\right)& =& 0\end{array}$

$\begin{array}{rcl}x& =& 4orx=-4\end{array}$

The intercepts on the x-axis are $\left(4,0\right)$and $\left(-4,0\right)$.