What is area y^2+8x=16,y^2-24x=48 ?

The answer to area y^2+8x=16,y^2-24x=48 is No Solution

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area y^2+8x=16,y^2-24x=48

area

y^{2} + 8 x = 16, y^{2} - 24 x = 48

No Solution

Solution steps

Isolate

y

for

y^{2} + 8 x = 16 : y = 22 2 - x, y = - 22 2 - x

Isolate

y

for

y^{2} - 24 x = 48 : y = 26 2 + x, y = - 26 2 + x

f_{1} (x) = 22 2 - x

f_{2} (x) = - 22 2 - x

To find the intersection points solve

f_{i} (x) = f_{j} (x)

22 2 - x = - 22 2 - x : x = 2

22 2 - x = 26 2 + x : x = - 1

22 2 - x = - 26 2 + x :

No Solution for

x \in R

- 22 2 - x = 26 2 + x :

No Solution for

x \in R

- 22 2 - x = - 26 2 + x : x = - 1

26 2 + x = - 26 2 + x : x = - 2

Curves not bounded

= No Solution

(8 x)^{2 x}

\frac{\partial}{\partial s} (\frac{s - 2 r}{t})

\int \frac{3 x + 1}{x} d x

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 0

\frac{d}{d x} (2 x^{2} + 7)

What is area y^2+8x=16,y^2-24x=48 ?
The answer to area y^2+8x=16,y^2-24x=48 is No Solution