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

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Presentation transcript:

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

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

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

Ellipse & hyperbola

How Many Points of Intersection? Hyperbola & line

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

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 = 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).

ANSWER The solution is (3, 1).

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 x – 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

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 y 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)

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 (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.

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)

Solve the system by elimination. 9x 2 + y 2 – 90x = 0 Equation 1 x 2 – y 2 – 16 = 0 Equation 2 SOLUTION 9x 2 + y 2 – 90x = 0 x 2 – y 2 – 16 = 0 10x 2 – 90x = 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.

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

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

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

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

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