What is 4(x+1)^2+3(y-1)^2=48 ?

The solution to 4(x+1)^2+3(y-1)^2=48 is Ellipse with (h,k)=(-1,1),a=2sqrt(3),b=4

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4(x+1)^2+3(y-1)^2=48

4 (x + 1)^{2} + 3 (y - 1)^{2} = 48

(h, k) = (- 1, 1), a = 23, b = 4

Solution steps

4 (x + 1)^{2} + 3 (y - 1)^{2} = 48

Rewrite

4 (x + 1)^{2} + 3 (y - 1)^{2} = 48

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{( 2 3 ) ^{2}} + \frac{( y - 1 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 1), a = 23, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 1), b = 4, a = 23

What is 4(x+1)^2+3(y-1)^2=48 ?
The solution to 4(x+1)^2+3(y-1)^2=48 is Ellipse with (h,k)=(-1,1),a=2sqrt(3),b=4