What is ((x-10)^2)/(100)-(y+3^2)/(36)=1 ?

The solution to ((x-10)^2)/(100)-(y+3^2)/(36)=1 is Parabola with (h,k)=(10,-45),p= 25/36

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((x-10)^2)/(100)-(y+3^2)/(36)=1

\frac{( x - 10 ) ^{2}}{100} - \frac{y + 3 ^{2}}{36} = 1

(h, k) = (10, - 45), p = \frac{25}{36}

Solution steps

\frac{( x - 10 ) ^{2}}{100} - \frac{y + 3 ^{2}}{36} = 1

Rewrite

\frac{( x - 10 ) ^{2}}{100} - \frac{y + 3 ^{2}}{36} = 1

in the standard form:

4 \cdot \frac{25}{36} (y - (- 45)) = (x - 10)^{2}

Therefore parabola properties are:

(h, k) = (10, - 45), p = \frac{25}{36}

x, \frac{104}{x} = \frac{500}{0.03 m ( x ) ^{2}}

x^{2} + 2 x + 4 y = 19

(x - 4)^{2} = 4 (- 6) (y - 2)

y^{2} - 6 y + 7 x - 7 = 0

x^{2} + 2 x - 3 = o

What is ((x-10)^2)/(100)-(y+3^2)/(36)=1 ?
The solution to ((x-10)^2)/(100)-(y+3^2)/(36)=1 is Parabola with (h,k)=(10,-45),p= 25/36