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.

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26-01-2020
Mathematics

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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.

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rani01654 rani01654 · Answer 1 · 2020-02-01T11:04:23+01:00

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 [tex]y = - \frac{1}{3}(x - 1)^{2} + 8[/tex] and y = x² - 2x - 3.

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

So, [tex]- \frac{1}{3}(x - 1)^{2} + 8 = x^{2} - 2x - 3[/tex]

⇒ - (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)

MrRoyal MrRoyal · Answer 2 · 2021-12-07T14:02:17+01:00

The area formed by the triangle is 15 square units

The functions of the parabolas are given as:

[tex]\mathbf{y = -\frac 13(x - 1)^2 + 8}[/tex]

[tex]\mathbf{y = x^2 - 2x - 3}[/tex]

Start by calculating the points of intersection, by equating both functions

[tex]\mathbf{-\frac 13(x - 1)^2 + 8 = x^2 - 2x - 3}[/tex]

Evaluate the exponent

[tex]\mathbf{-\frac 13(x^2 - 2x + 1) + 8 = x^2 - 2x - 3}[/tex]

Multiply through by -3

[tex]\mathbf{x^2 - 2x + 1 - 24 = -3x^2 + 6x + 9}[/tex]

[tex]\mathbf{x^2 - 2x - 23 = -3x^2 + 6x + 9}[/tex]

Collect like terms

[tex]\mathbf{3x^2 + x^2 - 2x - 6x -23-9 =0}[/tex]

[tex]\mathbf{4x^2 - 8x -32 =0}[/tex]

Divide through by 4

[tex]\mathbf{x^2 - 2x -8 =0}[/tex]

Expand

[tex]\mathbf{x^2 - 4x + 2x -8 =0}[/tex]

Factorize

[tex]\mathbf{x(x- 4) + 2(x -4) =0}[/tex]

Factor out x - 4

[tex]\mathbf{(x+ 2)(x -4) =0}[/tex]

Split

x = -2, or x = 4

Solve for y, when x = -2 and 4

[tex]\mathbf{y = x^2 - 2x - 3}[/tex]

[tex]\mathbf{y = (-2)^2 - 2(-2) - 3 = 5}[/tex]

[tex]\mathbf{y = (4)^2 - 2(4) - 3 = 5}[/tex]

So, the points of intersection are (-2,5) and (4,5)

Calculate the value of x.

[tex]\mathbf{x = |-2-4|}[/tex]

[tex]\mathbf{x = |-6|}[/tex]

Remove the absolute bracket

[tex]\mathbf{x = 6}[/tex]

So, we have:

[tex]\mathbf{x = 6}[/tex] and [tex]\mathbf{y = 5}[/tex]

The area is then calculated as:

[tex]\mathbf{Area = \frac 12xy}[/tex]

This gives

[tex]\mathbf{Area = \frac 12 \times 6 \times 5}[/tex]

[tex]\mathbf{Area = 15}[/tex]

Hence, the area formed by the triangle is 15 square units