4x^2+y^2<= 17

Lösung

4 x^{2} + y^{2} \leq 17

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

Schritte zur Lösung

4 x^{2} + y^{2} \leq 17

Rewrite

4 x^{2} + y^{2} \leq 17

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{( x + 8 ) ^{2}}{20} + \frac{( y + 36 ) ^{2}}{36} = 1

4 x^{2} + y^{2} - 16 x - 2 y + 1 = 0

5 x^{2} + z^{2} = 100

\frac{25 x ^{2}}{400 - 1 2 ^{2}} + \frac{16 y ^{2}}{400 - 1 2 ^{2}} = 1

\frac{( x - 2.1 ) ^{2}}{2.25} + \frac{( y + 1.6 ) ^{2}}{6.16} = 1