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The lower vertices of a square lie on the x-axis while the upper vertices lie on the parabola given by y = 35 − x^2 . What is the area of the square?

May 12, 2020

#1
+1

Let the two vertices on the x-axis be  (a,0)  and  (-a,0).

The length of the base will be  2a.

So, the vertex in Quadrant I, must have coordinates  (a,2a)  making  x = a  and  y = 2a.

For the parabola:  y =  35 - x2,  the point  (a, 2a)  becomes:  2a  =  35 - a2.

Solving this equation:  a2 - 2a - 35  =  0     --->      (a - 7)(a - 5)  =  0

If  a = 5, then  y = 35 - 52  =  35 - 25  =  10.

This makes the base 10 and the height 10, for an area of 100.

If  a = 7, then  y = 35 - 72  =  35 - 49  =  -14     <--- impossible for Quadrant I, but possible for Quadrant IV.

This makes the base 14 and the height 14, for an area of 196.

This square in not completly contained within the parabola, but it is a square.

It doesn't fulfill the requirements of the problem, which states that the lower vertices lie on the x-axis.

May 12, 2020

#1
+1

Let the two vertices on the x-axis be  (a,0)  and  (-a,0).

The length of the base will be  2a.

So, the vertex in Quadrant I, must have coordinates  (a,2a)  making  x = a  and  y = 2a.

For the parabola:  y =  35 - x2,  the point  (a, 2a)  becomes:  2a  =  35 - a2.

Solving this equation:  a2 - 2a - 35  =  0     --->      (a - 7)(a - 5)  =  0

If  a = 5, then  y = 35 - 52  =  35 - 25  =  10.

This makes the base 10 and the height 10, for an area of 100.

If  a = 7, then  y = 35 - 72  =  35 - 49  =  -14     <--- impossible for Quadrant I, but possible for Quadrant IV.

This makes the base 14 and the height 14, for an area of 196.

This square in not completly contained within the parabola, but it is a square.

It doesn't fulfill the requirements of the problem, which states that the lower vertices lie on the x-axis.

geno3141 May 12, 2020
#2
0

Excellent, geno  !!!!   CPhill  May 12, 2020