4x^2+2y^2-7x+y-5=0

解

4 x^{2} + 2 y^{2} - 7 x + y - 5 = 0

(h, k) = (\frac{7}{8}, - \frac{1}{4}), a = \frac{131}{8}, b = \frac{131}{4 2}

解答ステップ

4 x^{2} + 2 y^{2} - 7 x + y - 5 = 0

4 x^{2} + 2 y^{2} - 7 x + y - 5 = 0

を標準的な楕円のequationの形式で書き換える

\frac{( x - \frac{7}{8} ) ^{2}}{( \frac{131}{8} ) ^{2}} + \frac{( y - ( - \frac{1}{4} ) ) ^{2}}{( \frac{131}{4 2} ) ^{2}} = 1

ゆえに楕円の特性は:

(h, k) = (\frac{7}{8}, - \frac{1}{4}), a = \frac{131}{8}, b = \frac{131}{4 2}

b > a

ゆえに

b

は半長径,

a

は半短径である

中心が (h, k) = (\frac{7}{8}, - \frac{1}{4}), b = \frac{131}{4 2}, a = \frac{131}{8} の楕円