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

解

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

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

解答ステップ

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

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

を標準的な楕円のequationの形式で書き換える

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

ゆえに楕円の特性は:

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

b > a

ゆえに

b

は半長径,

a

は半短径である

中心が (h, k) = (\frac{2}{3}, - 2), b = \frac{2 2}{3}, a = \frac{2 2}{3} の楕円