Question

Find the values of c such that the area of the region bounded by the parabola

y = 16x2 − c2 and y = c2 − 16x2
is 250/3. (Enter your answers as a comma-separated list.)

Answer 1

To find the values of c such that the area of the region bounded by the parabolas is 250/3, set the equations equal to each other, simplify, and solve the quadratic equation.

Step-by-step explanation:

We want to find the values of c such that the area of the region bounded by the parabolas y = 16x^2 - c^2 and y = c^2 - 16x^2 is 250/3.

To find the area, we need to find the points of intersection of the two parabolas.

Setting the equations equal to each other, we have 16x^2 - c^2 = c^2 - 16x^2.

Simplifying this equation, we get 32x^2 = 2c^2. Substituting c^2 = 32x^2 into y = c^2 - 16x^2, we have y = 32x^2 - 16x^2 = 16x^2.

Now we can find the points of intersection by solving the quadratic equation 16x^2 - 250/3 = 0.

Solving for x, we get x = ±√(250/48). Substituting these values of x back into the equation for y, we can find the corresponding values of c.

Hence, the values of c are ±√(48/3) or ±4√2.

Answer 2

The two parabolas are,

`y=16x^2-c^2 \ and \ y=c^2-16x^2`

By solving the above two equations we calculate where the two parabolas meet,

`So \ 16x^2-c^2 = c^2-16x^2 \Rightarrow 32x^2=2c^2 \Rightarrow x=(1)/(4)c`

Given the symmetry, the area bounded by the two parabolas is twice the area bounded by either parabola with the x-axis.

`\therefore Area=2\int_(-c)^(c)y.dx= 2\int_(-c)^(c)(16x^2-c^2).dx\\=2[(16)/(3)x^3-c^2x]_(-c)^( \ c)=2[((16)/(3)c^3-c^3)-(-(16)/(3)c^3+c^3)]=2[(32)/(3)c^3-2c^3]=2((26c^3)/(3))\\=(52c^3)/(3)`

`So (52c^3)/(3)=(250)/(3)\Rightarrow c=\sqrt[3]{(250)/(52)}=1.68`