Question

Find the area of the triangle formed by the origin and the points of intersection of parabolas y=− 1/3 (x−1)^2+8 and y=x^2−2x−3.

Answer 1

Answer:

The area of the triangle formed by origin, and the points (4,5) and (-2,5) will be 15 sq. units.

Step-by-step explanation:

The two parabolas are $y = - \frac{1}{3}(x - 1)^{2} + 8$ and y = x² - 2x - 3.

Now, solving those two equations we will get the points of intersection.

So, $- \frac{1}{3}(x - 1)^{2} + 8 = x^{2} - 2x - 3$

⇒ - (x - 1)² + 24 = 3x² - 6x - 9

⇒ -x² + 2x - 1 + 24 = 3x² - 6x - 9

⇒ 4x² - 8x - 32 = 0

⇒ x² - 2x - 8 = 0

⇒ x² - 4x + 2x - 8 = 0

⇒ (x - 4)(x + 2) = 0

So, x = 4 or - 2.

Now, for x = 4 , y = 4² - 2(4) - 3 = 5 and the point of intersection is (4,5).

Or, for x = - 2, y = (- 2)² - 2(- 2) - 3 = 5 and the point of intersection is (-2,5).

Now, points (4,5) and (-2,5) make a straight line parallel to the x-axis at a perpendicular distance of 5 units from origin and its length is (4 - (- 2)) = 6 units.

So, the area of the triangle formed by origin, and the points (4,5) and (-2,5) will be = 0.5 × 5 × 6 = 15 sq. units. (Answer)