What is 10s-0.1s^2= 1/2 (v^2-5^2) ?

The solution to 10s-0.1s^2= 1/2 (v^2-5^2) is Ellipse with (h,k)=(50,0),a=5sqrt(105),b=5sqrt(21)

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10s-0.1s^2= 1/2 (v^2-5^2)

10 s - 0.1 s^{2} = \frac{1}{2} (v^{2} - 5^{2})

(h, k) = (50, 0), a = 5105, b = 521

Solution steps

10 s - 0.1 s^{2} = \frac{1}{2} (v^{2} - 5^{2})

Rewrite

10 s - 0.1 s^{2} - \frac{1}{2} (v^{2} - 5^{2}) = 0

in the form of the standard ellipse equation

\frac{( s - 50 ) ^{2}}{( 5 105 ) ^{2}} + \frac{( v - 0 ) ^{2}}{( 5 21 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (50, 0), a = 5105, b = 521

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (50, 0), a = 5105, b = 521

5 x^{2} + 6 y^{2} + 70 x + 84 y + 509 = 0

2 x^{2} + 4 y^{2} - 8 = 0

9 (x - 1)^{2} + 4 (y + 2)^{2} = 36

3 x^{2} + 5 y + 5 y^{2} = 5

\frac{( x - 6 ) ^{2}}{36} + \frac{( y + 4 ) ^{2}}{16} = 1

What is 10s-0.1s^2= 1/2 (v^2-5^2) ?
The solution to 10s-0.1s^2= 1/2 (v^2-5^2) is Ellipse with (h,k)=(50,0),a=5sqrt(105),b=5sqrt(21)