What is 2x^2+3y^2=17 ?

The solution to 2x^2+3y^2=17 is Ellipse with (h,k)=(0,0),a=sqrt(17/2),b=sqrt(17/3)

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2x^2+3y^2=17

2 x^{2} + 3 y^{2} = 17

(h, k) = (0, 0), a = \frac{17}{2}, b = \frac{17}{3}

Solution steps

2 x^{2} + 3 y^{2} = 17

Rewrite

2 x^{2} + 3 y^{2} = 17

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{17}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{17}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{17}{2}, b = \frac{17}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = \frac{17}{2}, b = \frac{17}{3}

x^{2} + 4 y^{2} \leq 1

\frac{( X - 5 ) ^{2}}{25} + \frac{( y + 7 ^{2} ) ^{2}}{16} = 1

- x^{2} + \frac{2 x}{21} - \frac{17}{21} = y^{2}

\frac{( x - ( - 0.31 ) ) ^{2}}{3.49} + \frac{y ^{2}}{3.71} = 2

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{36} = 1

What is 2x^2+3y^2=17 ?
The solution to 2x^2+3y^2=17 is Ellipse with (h,k)=(0,0),a=sqrt(17/2),b=sqrt(17/3)