I is a parabola with (1,-18) as its vertex and opening upwards touches (4,0) which is one of the roots , a straight line y=-x+11 pass through the parabola at P and Q , find the coordinates of P and Q.

Guest Feb 11, 2015

#3**+10 **

**I is a parabola with (1,-18) as its vertex and opening upwards touches (4,0) which is one of the roots , a straight line y=-x+11 pass through the parabola at P and Q , find the coordinates of P and Q.**

$$\boxed{Parabola: y=a(x-x_v)^2+y_v }\qquad x_v=1 \quad y_v=-18\\

a? \quad x = 4 \quad y = 0 \\

0=a(4-1)^2-18\\\\

a=\dfrac{18}{3^2}=\frac{18}{9}=2\\\\

\boxed{Parabola: y=2(x-1)^2-18 \quad line: y=-x+11}$$

The cut:

$$y=-x+11\\

y+x=11 \\

x = 11-y \\

x-1 = 10-y$$

We set x-1 = 10 - y in Parabola:

$$y=2(x-1)^2-18 \\

y=2(10-y)^2-18 \\

y=2(100-20y+y^2)-18\\

y=200-40y+2y^2-18\\

\boxed{2y^2-41y+182=0}\\

y_{1,2}=\dfrac{41 \pm \sqrt{41^2-4*2*182} }{2*2}=\dfrac{41 \pm \sqrt{225} }{4}=\dfrac{41 \pm 15 }{4}\\\\

y_1=\frac{56}{4}=14 \qquad y_2 = \frac{26}{4}=6.5\\

x_1 = 11-y_1 =11-14=-3 \qquad x_2=11-y_2=11-6.5=4.5$$

Point P = (-3, 14) and Point Q = (4.5, 6.5)

heureka
Feb 11, 2015

#1**+10 **

I is a parabola with (1,-18) as its vertex and opening upwards touches (4,0) which is one of the roots , a straight line y=-x+11 pass through the parabola at P and Q , find the coordinates of P and Q.

(x-1)^2=4a(y+18) Passes through (4,0)

(4-1)^2=4a(0+18)

9=72a

a=9/72=1/8

So the equation of the parabola is

(x-1)^2=4*(1/8)(y+18)

(x-1)^2=(1/2)(y+18)

2(x-1)^2=y+18

y=2(x-1)^2-18

y=2(x^2-2x+1)-18

y=2x^2-4x+2-18

y=2x^2-4x-16

Now we need to solve this simultaneously with y=-x+11

-x+11=2x^2-4x-16

2x^2-3x-27=0

I want two numbers that add to -3 and mult to 2*-27=2*3*-9=6*-9 The numbers are 6 and -9

replace -3x with 6x-9x

2x^2+6x-9x-27=0

2x(x+3)-9(x+3)=0

(2x-9)(x+3)=0

2x-9=0 or x+3=0

2x=9 or x=-3

x=4.5 or x=-3

Now we need to solve y=2x^2-4x-16 simultaneously with y=-x+11

When x=4.5 y=-4.5+11=6.5

check $${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{4.5}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{4.5}}{\mathtt{\,-\,}}{\mathtt{16}} = {\frac{{\mathtt{13}}}{{\mathtt{2}}}} = {\mathtt{6.5}}$$ well that one works

When x=-3 y=--3+11=3+11=14

check $${\mathtt{2}}{\mathtt{\,\times\,}}{\left(-{\mathtt{3}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}\left(-{\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{16}} = {\mathtt{14}}$$ good that one is correct too

so P(-3,14) and Q( 4.5, 6.5) [or vise versa]

Melody
Feb 11, 2015

#3**+10 **

Best Answer

$$\boxed{Parabola: y=a(x-x_v)^2+y_v }\qquad x_v=1 \quad y_v=-18\\

a? \quad x = 4 \quad y = 0 \\

0=a(4-1)^2-18\\\\

a=\dfrac{18}{3^2}=\frac{18}{9}=2\\\\

\boxed{Parabola: y=2(x-1)^2-18 \quad line: y=-x+11}$$

The cut:

$$y=-x+11\\

y+x=11 \\

x = 11-y \\

x-1 = 10-y$$

We set x-1 = 10 - y in Parabola:

$$y=2(x-1)^2-18 \\

y=2(10-y)^2-18 \\

y=2(100-20y+y^2)-18\\

y=200-40y+2y^2-18\\

\boxed{2y^2-41y+182=0}\\

y_{1,2}=\dfrac{41 \pm \sqrt{41^2-4*2*182} }{2*2}=\dfrac{41 \pm \sqrt{225} }{4}=\dfrac{41 \pm 15 }{4}\\\\

y_1=\frac{56}{4}=14 \qquad y_2 = \frac{26}{4}=6.5\\

x_1 = 11-y_1 =11-14=-3 \qquad x_2=11-y_2=11-6.5=4.5$$

Point P = (-3, 14) and Point Q = (4.5, 6.5)

heureka
Feb 11, 2015