Question

The  area of triangle formed by points of intersection of parabola y=a(x+5)(x−1) with the coordinate axes is 12. Find a if it is known that parabola opens upward.

Answer 1

We know that the points at which the parabola intersects the x axis are

(-5,0) and (1,0)

So the extent between these two points would be the base of the triangle

lets find the length of the base using the distance formula

$\sqrt{[(-5-1)^{2}+(0-0)^{2} ]}$

the base b=6

We will get the height of the triangle when we put x=0 in the equation

y=a(0+5)(0-1)

y=-5a

so height = -5a (we take +5a since it is the height)

We know that the area of the triangle =

$\frac{1}{2}$ × 6 × (5a) = 12

15a=12

a= $\frac{4}{5}$