This document is an assignment for an Analytic Geometry course (AM126) from the Papua New Guinea University of Technology. It contains six questions that focus on circles, parabolas, ellipses, and hyperbolas. I’ll provide an overview and solution for each problem:

QUESTION ONE [5 Marks]

Express in standard form the equation of the circle that’s centered at (5, -2) and has a radius of 2.

Solution: The standard form of a circle’s equation is:

$(x - h)^2 + (y - k)^2 = r^2$

where $(h, k)$ is the center and $r$ is the radius. Plugging in the values:

• Center: $(5, -2)$
• Radius: $2$

$(x - 5)^2 + (y + 2)^2 = 4$

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QUESTION TWO [5 Marks]

Find an equation(s) of the circle(s) containing $(1, -4)$ and the points of intersection of:

1. $x^2 + y^2 + 2x - 4y + 1 = 0$
2. $x^2 + y^2 + 4x + 6y - 3 = 0$

Solution: The points of intersection can be found by solving the two given equations simultaneously. Subtract equation 1 from equation 2:

$(4x + 6y - 3) - (2x - 4y + 1) = 0$ Simplifying:

$2x + 10y - 4 = 0 \quad \Rightarrow \quad x + 5y = 2 \quad \text{(Equation 3)}$

Solve Equation 3 with one of the original equations to find the points of intersection. Once the points are found, use these and $(1, -4)$ to determine the circle’s equation.

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QUESTION THREE [10 Marks]

Sketch and find the equation of the parabola having vertex $(0, 0)$, axis along the x-axis, and passing through $(2, -1)$.

Solution: For a parabola with its vertex at $(0,0)$ and axis along the x-axis, the standard form is:

$y^2 = 4px$

Use the point $(2, -1)$:

$(-1)^2 = 4p(2) \quad \Rightarrow \quad 1 = 8p \quad \Rightarrow \quad p = \frac{1}{8}$

Thus, the equation is:

$y^2 = \frac{1}{2}x$

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QUESTION FOUR [5 Marks]

Find the equation of the ellipse which has a minor axis of length 8 and a vertex at $(0, -5)$.

Solution:

• Minor axis length = 8 $\Rightarrow$ $2b = 8 \Rightarrow b = 4$
• Vertex at $(0, -5)$

If the center is at the origin and the vertex is on the y-axis, the standard form is:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

From the vertex:

$a = 5 \quad \Rightarrow \quad a^2 = 25$

$\frac{x^2}{25} + \frac{y^2}{16} = 1$

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QUESTION FIVE [5 Marks]

Sketch the ellipse:

$\frac{(x - 1)^2}{25} + \frac{(y + 2)^2}{9} = 1$

Solution:

• Center: $(1, -2)$
• $a^2 = 25 \Rightarrow a = 5$
• $b^2 = 9 \Rightarrow b = 3$

Sketch the ellipse using these parameters.

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QUESTION SIX [5 Marks]

Sketch the hyperbola:

$\frac{y^2}{25} - \frac{x^2}{4} = 1$

Solution:

• Center: $(0, 0)$
• $a^2 = 25 \Rightarrow a = 5$
• $b^2 = 4 \Rightarrow b = 2$

The hyperbola opens up and down because $y^2$ is positive.

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Let me know if you'd like detailed steps for any specific question!

Related Questions:

1. How do you find the intersection points of two circles algebraically?
2. What are the properties of parabolas with a horizontal axis?
3. How can you derive the standard form of an ellipse from its vertices and foci?
4. What is the procedure to graph a hyperbola?
5. How do transformations affect the graphs of conic sections?

Tip:

For circles and other conic sections, always identify the center and relevant distances (radius, vertices, etc.) before attempting to write the equation.