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# the coordinate of P and Q

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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
+18835
+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
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#1
+91481
+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
#2
+81077
+5

Nice, Melody...

I always like these kind of problems...

CPhill  Feb 11, 2015
#3
+18835
+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

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