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


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


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

$h = 5a$

$Area= \frac{1}{2} \times base \times height$

$12= \frac{1}{2} \times 6 \times 5a$

$12 = 15a$

$a = \frac{12}{15}$

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

Since

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