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8.7 Solve Quadratic Systems p. 534 How do you find the points of intersection of conics?

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Presentation on theme: "8.7 Solve Quadratic Systems p. 534 How do you find the points of intersection of conics?"— Presentation transcript:

1 8.7 Solve Quadratic Systems p. 534 How do you find the points of intersection of conics?

2 How Many Points of Intersection? Circle & line Circle and parabola

3 Circle & ellipse Circle & hyperbola How Many Points of Intersection?

4 Ellipse & hyperbola

5 How Many Points of Intersection? Hyperbola & line

6 Find the points of intersection of the graphs of x 2 + y 2 = 13 and y = x + 1. Left side: substitute x = 2 right side: x = −3 into one of the equations and solve for y. The points of intersection are (2,3) and (−2, −3). x 2 + y 2 = 13 x 2 + (x + 1) 2 = 13 x 2 + x 2 + 2x + 1 = 13 2x 2 + 2x − 12 = 0 2(x − 2)(x + 3) = 0 x = 2 or x = −3

7 Solve the system using substitution. x 2 + y 2 = 10 Equation 1 y = – 3x + 10 Equation 2 SOLUTION Substitute –3x + 10 for y in Equation 1 and solve for x. x 2 + y 2 = 10 x 2 + (– 3x + 10) 2 = 10 x 2 + 9x 2 – 60x + 100 = 10 10x 2 – 60x + 90 = 0 x 2 – 6x + 9 = 0 (x – 3) 2 = 0 x = 3 Equation 1 Substitute for y. Expand the power. Combine like terms. Divide each side by 10. Perfect square trinomial Zero product property y = – 3(3) + 10 = 1 To find the y-coordinate of the solution, substitute x = 3 in Equation 2. The solution is (3, 1).

8 ANSWER The solution is (3, 1).

9 5. y 2 – 2x – 10 = 0 y = x 1 – – SOLUTION Substitute – x – 1 for y in Equation 1 and solve for x. y 2 – 2x 2 – 10 = 0 (– x – 1) 2 – 2x – 10 = 0 x 2 + 1 + 2x – 2x – 10 = 0 x 2 – 9 = 0 x 2 = 9 Equation 1 Substitute for y. Expand the power. Combine like terms. Add 9 to each side. x = ±3 Simplify. To find the y-coordinate of the solution, substitute x = −3 and x = 3 in equation 2. y = −(–3) –1 = 2 y = −(3) –1 = −4 The solutions are (–3, 2), and (3, –4) ANSWER

10 Find the points of intersection of the graphs in the system. x 2 + 4y 2 − 4 = 0 (ellipse) −2y 2 + x + 2 = 0 (parabola) Solve for x x = 2y 2 − 2 Substitute (2y 2 − 2) 2 +4y 2 −4 = 0 4y 4 −8y 2 + 4 + 4y 2 − 4 = 0 4y 4 −4y 2 = 0 4y 2 (y 2 −1) = 0 4y 2 (y +1)(y −1) = 0 4y 2 = 0, y +1 = 0, y −1 = 0 y = 0, y = −1, y = 1 Left side: find x for y = −1 Right side: find x for y = 1 Solution: (−2, 0), (0, 1), (0, −1)

11 SOLUTION 4. y = 0.5x – 3 x 2 + 4y 2 – 4 = 0 Substitute 0.5x – 3 for y in Equation 2 and solve for x. x 2 + 4y 2 – 4 = 0 x 2 + 4 (0.5x – 3) 2 – 4 = 0 x 2 + y (0.25x 2 – 3x + 9) – 4 = 0 2x 2 – 12x + 32 = 0 x 2 – 6x + 16 = 0 Equation 2 Substitute for y. Expand the power. Combine like terms. Divide each side by 2. This equation has no solution.

12 Find the points of intersection of the graphs in the system. x 2 + y 2 −16x + 39 = 0 x 2 − y 2 −9 = 0 Eliminate y 2 by adding x 2 + y 2 −16x + 39 = 0 x 2 − y 2 −9 = 0 2x 2 −16x + 30 = 0 2(x 2 −8x + 15) = 0 2(x −5)(x −3) = 0 x = 3 or x = 5 Left: find y for x = 3 Right: find y for x = 5 Graphs intersect at: (3, 0), (5, 4), (5,−4)

13 Solve the system by elimination. 9x 2 + y 2 – 90x + 216 = 0 Equation 1 x 2 – y 2 – 16 = 0 Equation 2 SOLUTION 9x 2 + y 2 – 90x + 216 = 0 x 2 – y 2 – 16 = 0 10x 2 – 90x + 200 = 0 Add. x 2 – 9x + 20 = 0 Divide each side by 10. (x – 4)(x – 5) = 0 Factor x = 4 or x = 5 Zero product property Add the equations to eliminate the y 2 - term and obtain a quadratic equation in x. When x = 4, y = 0. When x = 5, y = ±3. ANSWER The solutions are (4, 0), (5, 3), and (5, 23), as shown.

14 Navigation A ship uses LORAN (long- distance radio navigation) to find its position.Radio signals from stations A and B locate the ship on the blue hyperbola, and signals from stations B and C locate the ship on the red hyperbola. The equations of the hyperbolas are given below. Find the ship’s position if it is east of the y - axis. x 2 – y 2 – 16x + 32 = 0 Equation 1 – x 2 + y 2 – 8y + 8 = 0 Equation 2

15 x 2 – y 2 – 16x + 32 = 0 Equation 1 – x 2 + y 2 – 8y + 8 = 0 Equation 2 SOLUTION STEP 1 Add the equations to eliminate the x 2 - and y 2 - terms. x 2 – y 2 – 16x + 32 = 0 – x 2 + y 2 – 8y + 8 = 0 – 16x – 8y + 40 = 0 Add. y = – 2x + 5 Solve for y. STEP 2 Substitute – 2x + 5 for y in Equation 1 and solve for x. x 2 – y 2 – 16x + 32 = 0 Equation 1 x 2 – (  2x + 5) 2 – 16x + 32 = 0 3x 2 – 4x – 7 = 0 Substitute for y. Simplify. (x + 1)(3x – 7) = 0 Factor. Zero product property x = – 1 or x = 7373

16 ANSWER Because the ship is east of the y - axis, it is at STEP 3

17 How do you find the points of intersection of conics? Use substitution or linear combination to solve for the point(s) of intersection

18 8-7Assignment Page 537, 9-15 odd, 23-27 (Quadratic formula will be helpful with #11)


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