(3p+2q)2=9p^2+4q^2

解

(3 p + 2 q) 2 = 9 p^{2} + 4 q^{2}

(h, k) = (\frac{1}{3}, \frac{1}{2}), a = \frac{2}{3}, b = \frac{1}{2}

解答ステップ

(3 p + 2 q) \cdot 2 = 9 p^{2} + 4 q^{2}

(3 p + 2 q) 2 - 9 p^{2} - 4 q^{2} = 0

を標準的な楕円のequationの形式で書き換える

\frac{( p - \frac{1}{3} ) ^{2}}{( \frac{2}{3} ) ^{2}} + \frac{( q - \frac{1}{2} ) ^{2}}{( \frac{1}{2} ) ^{2}} = 1

ゆえに楕円の特性は:

(h, k) = (\frac{1}{3}, \frac{1}{2}), a = \frac{2}{3}, b = \frac{1}{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{1}{3}, \frac{1}{2}), b = \frac{1}{2}, a = \frac{2}{3} の楕円