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

Answer:

$10\ \text{sq. units}$

Step-by-step explanation:

The parabola's are

$y=-x^2+9$

$y=2x^2-3$

So

$-x^2+9=2x^2-3\\\Rightarrow 12=3x^2\\\Rightarrow \\\Rightarrow x=\sqrt{\dfrac{12}{3}}\\\Rightarrow x=\pm 2$

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

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

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

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

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