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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.

 Feb 11, 2015

Best Answer 

 #3
avatar+26396 
+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.

Parabola:y=a(xxv)2+yvxv=1yv=18a?x=4y=00=a(41)218a=1832=189=2Parabola:y=2(x1)218line:y=x+11

The cut:

y=x+11y+x=11x=11yx1=10y

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

y=2(x1)218y=2(10y)218y=2(10020y+y2)18y=20040y+2y2182y241y+182=0y1,2=41±4124218222=41±2254=41±154y1=564=14y2=264=6.5x1=11y1=1114=3x2=11y2=116.5=4.5

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

 Feb 11, 2015
 #1
avatar+118696 
+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      2×4.524×4.516=132=6.5      well that one works

 

When x=-3       y=--3+11=3+11=14

check       2×(3)24×(3)16=14         good that one is correct too

 

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

 Feb 11, 2015
 #2
avatar+130458 
+5

Nice, Melody...

I always like these kind of problems...

 Feb 11, 2015
 #3
avatar+26396 
+10
Best Answer

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.

Parabola:y=a(xxv)2+yvxv=1yv=18a?x=4y=00=a(41)218a=1832=189=2Parabola:y=2(x1)218line:y=x+11

The cut:

y=x+11y+x=11x=11yx1=10y

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

y=2(x1)218y=2(10y)218y=2(10020y+y2)18y=20040y+2y2182y241y+182=0y1,2=41±4124218222=41±2254=41±154y1=564=14y2=264=6.5x1=11y1=1114=3x2=11y2=116.5=4.5

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

heureka Feb 11, 2015

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