12x-5y+10x^2+25=(-(100y^2))/(10+3)

Lösung

12 x - 5 y + 10 x^{2} + 25 = \frac{- ( 100 y ^{2} )}{10 + 3}

(h, k) = (- \frac{3}{5}, \frac{13}{40}), a = i \frac{3554}{40}, b = \frac{23101}{40 5}

Schritte zur Lösung

12 x - 5 y + 10 x^{2} + 25 = \frac{- 100 y ^{2}}{10 + 3}

Rewrite

12 x - 5 y + 10 x^{2} + 25 = \frac{- 100 y ^{2}}{10 + 3}

in the form of the standard ellipse equation

\frac{( x - ( - \frac{3}{5} ) ) ^{2}}{( i \frac{3554}{40} ) ^{2}} + \frac{( y - \frac{13}{40} ) ^{2}}{( \frac{23101}{40 5} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{3}{5}, \frac{13}{40}), a = i \frac{3554}{40}, b = \frac{23101}{40 5}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- \frac{3}{5}, \frac{13}{40}), a = i \frac{3554}{40}, b = \frac{23101}{40 5}

(16 - x^{2} - 9 y^{2}) = - 1

36 x^{2} + 17 y^{2} - 648 x + 238 y + 689 = 0

25 x^{2} + 16 y^{2} - 200 x - 160 y + 400 = 0

\frac{x ^{2}}{64} + \frac{( y - 6 ) ^{2}}{121} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{40} = 1