Question

find the area of the isocele triangle formed by the points of intersection of parabolas y=-x^2+9 and y=2x^2-3 and the origin

Answer 1

`10\ \text{sq. units}`

Explanation:

The parabola's are

`y=-x^2+9`

`y=2x^2-3`

So

`-x^2+9=2x^2-3\\\Rightarrow 12=3x^2\\\Rightarrow \\\Rightarrow x=\sqrt{(12)/(3)}\\\Rightarrow x=\pm 2`

`y=-x^2+9=-(2)^2+9\\\Rightarrow y=5`

`y=-(-2)^2+9=5`

So, the points at which the parabola's intersect each other is
`(2,5)` and
`(-2,5)`

The three points of the triangle are
`(2,5)`,
`(-2,5)` and
`(0,0)`

Area of a triangle is given by

`A=(1)/(2)[x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)]\\\Rightarrow A=(1)/(2)[2(5-0)+(-2)(0-5)+0(5-5)]\\\Rightarrow A=10\ \text{sq. units}`

Area of the triangle formed is
`10\ \text{sq. units}`.