10x^2+4y^2+2x+16y=144

Lösung

10 x^{2} + 4 y^{2} + 2 x + 16 y = 144

(h, k) = (- \frac{1}{10}, - 2), a = \frac{1601}{10}, b = \frac{1601}{2 10}

Schritte zur Lösung

10 x^{2} + 4 y^{2} + 2 x + 16 y = 144

Rewrite

10 x^{2} + 4 y^{2} + 2 x + 16 y = 144

in the form of the standard ellipse equation

\frac{( x - ( - \frac{1}{10} ) ) ^{2}}{( \frac{1601}{10} ) ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{( \frac{1601}{2 10} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{1}{10}, - 2), a = \frac{1601}{10}, b = \frac{1601}{2 10}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{1}{10}, - 2), b = \frac{1601}{2 10}, a = \frac{1601}{10}

y^{2} = - 36 x

x^{2} = 4 y

v er t i ces f (x) = x^{2} - 9

y^{2} = x - 3

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{9} = 1