Question

given the equations, identify which way the parabola opens by matching an equation on the left with a term on the right.
$x^2 = -10y$
$y^2 = 6x$
$x^2 = 3y$
$y^2 = -2x$
left
right
down
up

Explanation:

Step1: Recall parabola opening rules

For a parabola in the form \(x^2 = 4py\), it opens up if \(p>0\) and down if \(p < 0\). For a parabola in the form \(y^2=4px\), it opens right if \(p>0\) and left if \(p < 0\).

Step2: Analyze \(x^2=-10y\)

Rewrite as \(x^2 = 4py\) where \(4p=-10\), so \(p=-\frac{10}{4}=-\frac{5}{2}<0\). Thus, it opens down.

Step3: Analyze \(y^2 = 6x\)

Rewrite as \(y^2=4px\) where \(4p = 6\), so \(p=\frac{6}{4}=\frac{3}{2}>0\). Thus, it opens right.

Step4: Analyze \(x^2=3y\)

Rewrite as \(x^2=4py\) where \(4p = 3\), so \(p=\frac{3}{4}>0\). Thus, it opens up.

Step5: Analyze \(y^2=-2x\)

Rewrite as \(y^2=4px\) where \(4p=-2\), so \(p=-\frac{2}{4}=-\frac{1}{2}<0\). Thus, it opens left.

Answer:

• \(x^2=-10y\) matches down
• \(y^2 = 6x\) matches right
• \(x^2=3y\) matches up
• \(y^2=-2x\) matches left