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Rectangle ABCD is symmetric with respect to y-axis. Points A and B belong to the parabola y=x2. Points C and D are on the parabola y=−3x2+k. Find k if the area of the rectangle is 66 and length of AB is 6.

 Apr 8, 2017
 #1
avatar+9466 
+2

AB will be 6 when x = ± 3, that is when y = 9...

I don't have time to do any more but maybe this helps get you started.

 

Here's a rough drawing (Not to scale):

 

 Apr 8, 2017
 #2
avatar+128408 
+2

 

To continue where hectictar left off......

 

If AB  = 6   then, because we have symmetry with the y axis,  B must be the point (3, y)

 

And y  can be found as   y = 3^2   = 9

 

So B  = (3, 9)

 

And because the area of the rectangle is 66  and AB = 6, then BC  =  11

 

Then, C will have the coordinates (3, -2)

 

So ......using  point C and  y = -3x^2 + k....we can find k as:

 

-2  =  -3(3)^2   + k

-2  = -27  + k

25 =   k

 

Here's a pic :

 

 

cool cool cool

 Apr 8, 2017
edited by CPhill  Apr 8, 2017
edited by CPhill  Apr 8, 2017

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