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EXAMPLE 1

Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius. r = 36 = 6. EXAMPLE 1. Graph an equation of a circle. Graph y 2 = – x 2 + 36 . Identify the radius of the circle. SOLUTION. STEP 1.

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EXAMPLE 1

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  1. Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius r= 36 = 6. EXAMPLE 1 Graph an equation of a circle Graphy2 = – x2 + 36. Identify the radius of the circle. SOLUTION STEP 1 Rewrite the equationy2 = – x2 + 36in standard form asx2 + y2 = 36. STEP 2

  2. EXAMPLE 1 Graph an equation of a circle STEP 3 Draw the circle. First plot several convenient points that are 6 units from the origin, such as (0, 6), (6, 0), (0, –6), and (–6, 0). Then draw the circle that passes through the points.

  3. The radius is = 29 r = (2 –0)2+ (–5 –0)2 = 4 + 25 29 EXAMPLE 2 Write an equation of a circle The point (2, –5) lies on a circle whose center is the origin. Write the standard form of the equation of the circle. SOLUTION Because the point (2, –5) lies on the circle, the circle’s radius rmust be the distance between the center (0, 0) and (2, –5). Use the distance formula.

  4. Use the standard form withr to write an equation of the circle. =29 = (29 )2 Substitute for r 29 EXAMPLE 2 Write an equation of a circle x2 + y2 = r2 Standard form x2 + y2 x2 + y2 = 29 Simplify

  5. A line tangent to a circle is perpendicular to the radius at the point of tangency. Because the radius to the point (1–3, 2)has slope = = – 2 – 0 2 – 3 – 0 3 EXAMPLE 3 Standardized Test Practice SOLUTION m

  6. y –2= (x –(–3)) y –2= x + y = x + 13 2 ANSWER 3 3 3 3 9 2 2 2 2 2 The correct answer is C. EXAMPLE 3 Standardized Test Practice the slope of the tangent line at (23, 2) is the negative reciprocal of or An equation of 2 – 3 the tangent line is as follows: Point-slope form Distributive property Solve for y.

  7. In the diagram above, the origin represents the tower and the positive y-axis represents north. EXAMPLE 4 Write a circular model Cell Phones A cellular phone tower services a 10 mile radius. You get a flat tire 4 miles east and 9 miles north of the tower. Are you in the tower’s range? SOLUTION STEP 1 Write an inequality for the region covered by the tower. From the diagram, this region is all points that satisfy the following inequality: x2 + y2 < 102

  8. ? 42 + 92 < 102  97 < 100 ANSWER So, you are in the tower’s range. EXAMPLE 4 Write a circular model STEP 2 Substitute the coordinates (4, 9) into the inequality from Step 1. x2 + y2 < 102 Inequality from Step 1 Substitute for xandy. The inequality is true.

  9. EXAMPLE 5 Apply a circular model Cell Phones In Example 4, suppose that you fix your tire and then drive south. For how many more miles will you be in range of the tower ? SOLUTION When you leave the tower’s range, you will be at a point on the circle x2 + y2 = 102 whose x-coordinate is 4 and whose y-coordinate is negative. Find the point (4, y) where y < 0 on the circle x2 + y2 = 102.

  10. 84 + y = y +9.2 ANSWER Becausey < 0, y – 9.2. You will be in the tower’s range from (4, 9) to (4, –9.2), a distance of | 9 – (– 9.2) | = 18.2miles. EXAMPLE 5 Apply a circular model x2+ y2 = 102 Equation of the circle 42+ y2 = 102 Substitute 4 forx. Solve for y. Use a calculator.

  11. 100 x2 4x2 25x2 100 25 100 100 = + y24 + = 1 EXAMPLE 1 Graph an equation of an ellipse Graph the equation 4x2 + 25y2 = 100. Identify the vertices, co-vertices, and foci of the ellipse. SOLUTION STEP 1 Rewrite the equation in standard form. 4x2 + 25y2 = 100 Write original equation. Divide each side by 100. Simplify.

  12. soc = 21 The foci are at( + 21 , 0),or about ( + 4.6, 0). EXAMPLE 1 Graph an equation of an ellipse STEP 2 Identify the vertices, co-vertices, and foci. Note that a2 = 25 and b2 = 4, so a = 5and b = 2. The denominator of the x2 - term is greater than that of the y2 - term, so the major axis is horizontal. The vertices of the ellipse are at (+a, 0) = (+5, 0). The co-vertices are at (0, +b) = (0, +2). Find the foci. c2= a2– b2= 52– 22= 21,

  13. EXAMPLE 1 Graph an equation of an ellipse STEP 3 Draw the ellipse that passes through each vertex and co-vertex.

  14. EXAMPLE 2 Write an equation given a vertex and a co-vertex Write an equation of the ellipse that has a vertex at (0, 4), a co-vertex at (– 3, 0), and center at (0, 0). SOLUTION Sketch the ellipse as a check for your final equation. By symmetry, the ellipse must also have a vertex at (0, – 4) and a co-vertex at (3, 0). Because the vertex is on the y - axis and the co-vertex is on the x - axis, the major axis is vertical with a = 4, and the minor axis is horizontal with b = 3.

  15. + + x2 32 y2 42 x2 9 y2 16 = 1 EXAMPLE 2 Write an equation given a vertex and a co-vertex ANSWER = 1 An equation is or

  16. EXAMPLE 3 Solve a multi-step problem Lightning When lightning strikes, an elliptical region where the strike most likely hit can often be identified. Suppose it is determined that there is a 50% chance that a lightning strike hit within the elliptical region shown in the diagram. • Write an equation of the ellipse. • The area Aof an ellipse is A = π ab. Find the area of the elliptical region.

  17. The major axis is horizontal, with a = = 200 400 2 200 2 andb = = 100 y2 1002 x2 40,000 y2 10,000 x2 2002 or = 1 = 1 + An equation is + The area isA = π(200)(100) 62,800 square meters. EXAMPLE 3 Solve a multi-step problem SOLUTION STEP 1 STEP 2

  18. EXAMPLE 4 Write an equation given a vertex and a focus Write an equation of the ellipse that has a vertex at (– 8, 0), a focus at (4, 0), and center at (0, 0). SOLUTION Make a sketch of the ellipse. Because the given vertex and focus lie on the x - axis, the major axis is horizontal, with a = 8and c = 4. To find b, use the equation c2 = a2 – b2. 42 = 82 – b2 b2 = 82 – 42 = 48

  19. b = or 48, 3 4 + ANSWER y2 48 x2 64 y2 x2 82 = 1 = 1 An equation is or + 3,)2 (4 EXAMPLE 4 Write an equation given a vertex and a focus

  20. Note thatA = 4, B = 0,andC = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(4)(1) = – 16 BecauseB2– 4AC < 0andA = C, the conic is an ellipse. To graph the ellipse, first complete the square in x. EXAMPLE 6 Classify a conic Classify the conic given by4x2 + y2 – 8x – 8 = 0. Then graph the equation. SOLUTION 4x2 + y2 – 8x – 8 = 0 (4x2 – 8x) + y2 = 8 4(x2 – 2x) + y2 = 8 4(x2– 2x + ? ) + y2 = 8 + 4( ? )

  21. (x – 1)2 y2 12 + = 1 3 From the equation, you can see that(h, k) = (1, 0), a = 12 = 2 3 ,andb = 3. Use these facts to draw the ellipse. EXAMPLE 6 Classify a conic 4(x2– 2x + 1) + y2 = 8 + 4(1) 4(x – 1)2 + y2 = 12

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