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

ANSWER



`a = (4)/(5)`


Step-by-step explanation

The given parabola has equation,


`y = a(x + 5)(x - 1)`

The y-intercept is


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


`y = - 5a`

This gives one vertex of the triangle to be,


`(0,-5a)`

The x-intercept is


`a(x + 5)(x - 1) = 0`

This implies


`(x + 5)(x - 1) = 0`


`(x + 5) = 0 \: or \: (x - 1) = 0`


`x = - 5 \: or \: x = 1`

The other two vertices of the triangle is,


`(-5,0),(1,0)`

The height of this triangle is


`h = 5a`


`Area= (1)/(2) * base * height`


`12= (1)/(2) * 6 * 5a`


`12 = 15a`


`a = (12)/(15)`


`a = (4)/(5)`

Since


`a > \: 0`
it means the parabola opens upwards.