Question

lauren describes a parabola where the focus has a positive, nonzero x coordinate. which parabola(s) could lauren be describing? check all that apply. $x^2 = 4y$ $x^2 = -6y$ $y^2 = x$ $y^2 = 10x$ $y^2 = -3x$ $y^2 = 5x$

Explanation:

To determine the focus of each parabola, we use the standard forms of parabolas:

For parabolas of the form \( x^2 = 4py \) (opens up or down):

• The focus is at \( (0, p) \). So the \( x \)-coordinate of the focus is \( 0 \), which does not meet the requirement of a positive, nonzero \( x \)-coordinate.
• For \( x^2 = 4y \), comparing with \( x^2 = 4py \), we have \( 4p = 4 \implies p = 1 \). Focus: \( (0, 1) \) (x-coordinate 0, so exclude).
• For \( x^2 = -6y \), comparing with \( x^2 = 4py \), we have \( 4p = -6 \implies p = -\frac{3}{2} \). Focus: \( (0, -\frac{3}{2}) \) (x-coordinate 0, so exclude).

For parabolas of the form \( y^2 = 4px \) (opens right or left):

• The focus is at \( (p, 0) \). We need \( p > 0 \) (so that the \( x \)-coordinate \( p \) is positive and nonzero).
• For \( y^2 = x \), rewrite as \( y^2 = 4 \times \frac{1}{4}x \). So \( 4p = 1 \implies p = \frac{1}{4} \). Focus: \( (\frac{1}{4}, 0) \) (x-coordinate positive, include).
• For \( y^2 = 10x \), rewrite as \( y^2 = 4 \times \frac{5}{2}x \). So \( 4p = 10 \implies p = \frac{5}{2} \). Focus: \( (\frac{5}{2}, 0) \) (x-coordinate positive, include).
• For \( y^2 = -3x \), rewrite as \( y^2 = 4 \times (-\frac{3}{4})x \). So \( 4p = -3 \implies p = -\frac{3}{4} \). Focus: \( (-\frac{3}{4}, 0) \) (x-coordinate negative, exclude).
• For \( y^2 = 5x \), rewrite as \( y^2 = 4 \times \frac{5}{4}x \). So \( 4p = 5 \implies p = \frac{5}{4} \). Focus: \( (\frac{5}{4}, 0) \) (x-coordinate positive, include).

Answer:

• \( y^2 = x \)
• \( y^2 = 10x \)
• \( y^2 = 5x \)