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An equilateral triangle has all three of its vertices on the parabola $y=x^2-8+5$. One vertex of the triangle is on the vertex of the parabola, and the opposite side lies along the line $y=k$. What is the value of $k$?

 Jul 10, 2019
 #1
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I'm assuming that   the equation of the parabola is  y = x^2 - 8x + 5      (1)

 

The x coordinate of the vertex  is given by  - (-8) / 2  =  8 /2  = 4

And the y coordinate of the vertex is (4)^2 - 8(4) + 5  =  16 - 32 + 5  =  -11

 

One of the sides of the triangle  has one of its endpoints on the vertex and the other on the parabola ....and the line on which this side lies will form a 60° angle with the line y = -11...so...its slope  = tan 60° = √3

 

So....the equation of this line is :

y = √3 ( x - 4)  - 11     

y + 11 = √3 (x - 4)

 

[y + 11]

______   =  x   -  4

    √3

 

[ y + 11 ]

________ + 4      =  x

     √3

 

Sub this into (1)  for x  to   find  y = k

 

          [ y + 11]^2                [ y + 11]                      [y + 11 ]        

y =     _________    +  8    ________ + 16  -  (8) _______  - 32   + 5

                3                           √3                              √3   

 

y  =   [y^2 + 22y + 121]

       _______________   - 11

                     3  

 

3y = y^2 + 22y + 121 -33

 

y^2 + 19y  + 88  =  0        factor

 

(y + 11)  ( y + 8)  = 0

 

Set each factor to 0  and solve for y

 

y = - 11     (reject)

 

y = - 8    (accept)

 

So....k  = -8

 

Here's a pic :

 

 

 

cool cool cool

 Jul 10, 2019
edited by CPhill  Jul 11, 2019
edited by CPhill  Jul 13, 2019

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