What is (x-3y)/(x^2+y^2)+1=0 ?

The solution to (x-3y)/(x^2+y^2)+1=0 is Ellipse with (h,k)=(-1/2 , 3/2),a=sqrt(5/2),b=sqrt(5/2)

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(x-3y)/(x^2+y^2)+1=0

\frac{x - 3 y}{x ^{2} + y ^{2}} + 1 = 0

(h, k) = (- \frac{1}{2}, \frac{3}{2}), a = \frac{5}{2}, b = \frac{5}{2}

Solution steps

\frac{x - 3 y}{x ^{2} + y ^{2}} + 1 = 0

Rewrite

\frac{x - 3 y}{x ^{2} + y ^{2}} + 1 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- \frac{1}{2}, \frac{3}{2}), a = \frac{5}{2}, b = \frac{5}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{1}{2}, \frac{3}{2}), b = \frac{5}{2}, a = \frac{5}{2}

a^{2} + \frac{a ^{2}}{2} + \frac{b ^{2}}{2} = 26

x^{2} + y^{2} = - 12 y

\frac{( x - 4 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

\frac{4 x}{x ^{2} + y ^{2}} = 2

\frac{4 x}{x ^{2} + y ^{2}} = 1

What is (x-3y)/(x^2+y^2)+1=0 ?
The solution to (x-3y)/(x^2+y^2)+1=0 is Ellipse with (h,k)=(-1/2 , 3/2),a=sqrt(5/2),b=sqrt(5/2)