Question

PLEASE HELP!!

Find the equation of a parabola with vertex on the y-axis if points P(x1, 0) and R(3, 9) belong the parabola, the area of △PQR is 40.5sq.units, and point Q has coordinates (3, 0).

Answer 1

Answer:

$y=-\dfrac{9}{40}x^2+\dfrac{441}{40}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}$

Step-by-step explanation:

If a parabola has its vertex on the y-axis, then its equation is

$y=ax^2+b$

This parabola passes through the point R(3,9), then

$9=a\cdot 3^2+b\\ \\9=9a+b$

The area of the right triangle PQR is

$A_{PQR}=\dfrac{1}{2}\cdot PQ\cdot QR$

Find PQ and QR, if $P(x_1,0),\ Q(3,0),\ R(3,9):$

$PQ=\sqrt{(x_1-3)^2+(0-0)^2}=|x_1-3|\\\\QR=\sqrt{(3-3)^2+(0-9)^2}=9$

Now,

$40.5=\dfrac{1}{2}\cdot |x_1-3|\cdot 9\\ \\90=9|x_1-3|\\ \\|x_1-3|=10\\ \\x_1-3=10\ \text{or}\ x_1-3=-10\\ \\x_1=13\ \text{or }x_1=-7$

We get two possible points $P_1(-7,0)$ and $P_2(13,0).$

For point $P_1:$

$0=a\cdot (-7)^2+b\\ \\49a+b=0$

So,

$b=-49a\\ \\9=9a-49a\\ \\-40a=9\\ \\a=-\dfrac{9}{40}\\ \\b=\dfrac{441}{40}\\ \\y=-\dfrac{9}{40}x^2+\dfrac{441}{40}$

For point $P_2:$

$0=a\cdot (13)^2+b\\ \\169a+b=0$

So,

$b=-169a\\ \\9=9a-169a\\ \\-160a=9\\ \\a=-\dfrac{9}{160}\\ \\b=\dfrac{1,521}{160}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}$