Question

determine if $16x^2 + 4y^2 = 1$ is an ellipse.
\bigcirc not an ellipse
\bigcirc ellipse, we can write the equation of the ellipse in standard form $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.
where:
$h = \square$
$k = \square$
$a = \square$
$b = \square$

Explanation:

Step1: Write the equation in standard form

Divide the entire equation \(16x^{2}+4y^{2}=1\) by \(1\). We get \(\frac{x^{2}}{\frac{1}{16}}+\frac{y^{2}}{\frac{1}{4}} = 1\). This is of the form \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\) where \(h = 0,k = 0\).

Step2: Identify \(a\) and \(b\)

If \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\), then for \(\frac{x^{2}}{\frac{1}{16}}+\frac{y^{2}}{\frac{1}{4}} = 1\), we have \(a^{2}=\frac{1}{16}\), so \(a=\frac{1}{4}\) and \(b^{2}=\frac{1}{4}\), so \(b = \frac{1}{2}\)

Answer:

Ellipse, we can write the equation of the ellipse in standard form \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\).
Where:
\(h = 0\)
\(k = 0\)
\(a=\frac{1}{4}\)
\(b=\frac{1}{2}\)