What is ((x+3)^2)/(24)+((y-5)^2)/(49)=1 ?

The solution to ((x+3)^2)/(24)+((y-5)^2)/(49)=1 is Ellipse with (h,k)=(-3,5),a=2sqrt(6),b=7

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((x+3)^2)/(24)+((y-5)^2)/(49)=1

\frac{( x + 3 ) ^{2}}{24} + \frac{( y - 5 ) ^{2}}{49} = 1

(h, k) = (- 3, 5), a = 26, b = 7

Solution steps

\frac{( x + 3 ) ^{2}}{24} + \frac{( y - 5 ) ^{2}}{49} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{24} + \frac{( y - 5 ) ^{2}}{49} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{( 2 6 ) ^{2}} + \frac{( y - 5 ) ^{2}}{7 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 5), a = 26, b = 7

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 5), b = 7, a = 26

\frac{( x + 3 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{100} = 1

y^{2} = 1 - 0.25 x^{2}

16 - x^{2} - 9 y^{2} = 0

256 x^{2} - 800 x + 625 = 400 x^{2} - 800 x + 400 y^{2} + 400

- 2 = \frac{2 x}{x ^{2} + y ^{2}}

What is ((x+3)^2)/(24)+((y-5)^2)/(49)=1 ?
The solution to ((x+3)^2)/(24)+((y-5)^2)/(49)=1 is Ellipse with (h,k)=(-3,5),a=2sqrt(6),b=7