What is ((x-8)^2)/(42)+((y+7)^2)/(16)=1 ?

The solution to ((x-8)^2)/(42)+((y+7)^2)/(16)=1 is Ellipse with (h,k)=(8,-7),a=sqrt(42),b=4

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((x-8)^2)/(42)+((y+7)^2)/(16)=1

\frac{( x - 8 ) ^{2}}{42} + \frac{( y + 7 ) ^{2}}{16} = 1

(h, k) = (8, - 7), a = 42, b = 4

Solution steps

\frac{( x - 8 ) ^{2}}{42} + \frac{( y + 7 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 8 ) ^{2}}{42} + \frac{( y + 7 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 8 ) ^{2}}{( 42 ) ^{2}} + \frac{( y - ( - 7 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (8, - 7), a = 42, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (8, - 7), a = 42, b = 4

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

6 x^{2} + 9 y^{2} + 54 y = 60 x - 87

16 x^{2} + 16 y^{2} = 1 + 2 x + x^{2}

(\frac{x ^{2}}{9}) + (\frac{y ^{2}}{5}) = 1

9 x^{2} + 25 y^{2} + 36 x + 150 y + 36 = 0

What is ((x-8)^2)/(42)+((y+7)^2)/(16)=1 ?
The solution to ((x-8)^2)/(42)+((y+7)^2)/(16)=1 is Ellipse with (h,k)=(8,-7),a=sqrt(42),b=4