Let's solve these intercepts step by step for both equations.

(a) $9x^2 + 16y^2 = 144$

1. Find the x-intercepts:

To find the x-intercepts, set $y = 0$ and solve for $x$.

$9x^2 + 16(0)^2 = 144$ $9x^2 = 144$ $x^2 = \frac{144}{9} = 16$ $x = \pm \sqrt{16} = \pm 4$

Thus, the x-intercepts are at $(4, 0)$ and $(-4, 0)$.

2. Find the y-intercepts:

To find the y-intercepts, set $x = 0$ and solve for $y$.

$9(0)^2 + 16y^2 = 144$ $16y^2 = 144$ $y^2 = \frac{144}{16} = 9$ $y = \pm \sqrt{9} = \pm 3$

Thus, the y-intercepts are at $(0, 3)$ and $(0, -3)$.

Summary for (a):

• x-intercept (smaller x-value): $(-4, 0)$
• x-intercept (larger x-value): $(4, 0)$
• y-intercept (smaller y-value): $(0, -3)$
• y-intercept (larger y-value): $(0, 3)$

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(b) $x^2 - xy + 3y = 64$

1. Find the x-intercepts:

To find the x-intercepts, set $y = 0$ and solve for $x$.

$x^2 - x(0) + 3(0) = 64$ $x^2 = 64$ $x = \pm \sqrt{64} = \pm 8$

Thus, the x-intercepts are at $(8, 0)$ and $(-8, 0)$.

2. Find the y-intercept:

To find the y-intercept, set $x = 0$ and solve for $y$.

$(0)^2 - (0)y + 3y = 64$ $3y = 64$ $y = \frac{64}{3} = \frac{64}{3}$

Thus, the y-intercept is at $\left(0, \frac{64}{3}\right)$.

Summary for (b):

• x-intercept (smaller x-value): $(-8, 0)$
• x-intercept (larger x-value): $(8, 0)$
• y-intercept: $\left(0, \frac{64}{3}\right)$

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Do you need more details or have any questions?

Here are 5 related questions:

1. How would you graph the equation of a circle like the one in part (a)?
2. What is the general form of the equation for an ellipse?
3. How can you solve for intercepts if an equation contains higher-order terms?
4. Can there be cases where there are no x- or y-intercepts? Why?
5. How do you determine the center and radii of an ellipse given an equation?

Tip: When finding intercepts, setting one variable to zero simplifies the equation, leading to a direct solution.