双曲線上の点 A A A α \alpha α 0 < α < π 2 \displaystyle 0<\alpha<\frac\pi2 0 < α < 2 π A = ( r cos α , r sin α ) A=(r\cos\alpha,r\sin\alpha) A = ( r cos α , r sin α ) x y = 1 xy=1 x y = 1 d y d x = − 1 x 2 = − y x = − tan α \displaystyle
\frac{dy}{dx}=-\frac1{x^2}=-\frac yx=-\tan\alpha d x d y = − x 2 1 = − x y = − tan α O A OA O A min ( 2 α , π − 2 α ) \displaystyle \min(2\alpha,\pi-2\alpha) min ( 2 α , π − 2 α ) π / 6 \pi/6 π /6 α = π / 12 \alpha=\pi/12 α = π /12 5 π / 12 5\pi/12 5 π /12 x y = r 2 2 sin 2 α = 1 \displaystyle
xy=\frac{r^2}{2}\sin2\alpha=1 x y = 2 r 2 sin 2 α = 1 r 2 = 2 / sin 2 α = 4 r^2=2/\sin2\alpha=4 r 2 = 2/ sin 2 α = 4 C 2 C_2 C 2 x 2 + y 2 = 4 x^2+y^2=4 x 2 + y 2 = 4
円と双曲線の交点の x x x x 2 + 1 x 2 = 4 \displaystyle
x^2+\frac1{x^2}=4 x 2 + x 2 1 = 4 x > 0 x>0 x > 0 x 4 − 4 x 2 + 1 = 0 x^4-4x^2+1=0 x 4 − 4 x 2 + 1 = 0 x 2 = 2 ± 3 x^2=2\pm\sqrt3 x 2 = 2 ± 3 p = 2 − 3 , q = 2 + 3 ( 0 < p < q , p q = 1 ) . p=\sqrt{2-\sqrt3},\qquad q=\sqrt{2+\sqrt3}\quad(0<p<q,\ pq=1). p = 2 − 3 , q = 2 + 3 ( 0 < p < q , pq = 1 ) . p ≦ x ≦ q p\le x\le q p ≦ x ≦ q F ( x ) = 4 − x 2 − 1 / x F(x)=\sqrt{4-x^2}-1/x F ( x ) = 4 − x 2 − 1/ x F ( 1 ) = 3 − 1 > 0 F(1)=\sqrt3-1>0 F ( 1 ) = 3 − 1 > 0
S = ∫ p q 4 − x 2 d x − ln q p . \displaystyle
S=\int_p^q\sqrt{4-x^2}\,dx-\ln\frac qp. S = ∫ p q 4 − x 2 d x − ln p q . p / 2 = sin ( π / 12 ) p/2=\sin(\pi/12) p /2 = sin ( π /12 ) q / 2 = sin ( 5 π / 12 ) q/2=\sin(5\pi/12) q /2 = sin ( 5 π /12 ) x = 2 sin u x=2\sin u x = 2 sin u π / 12 ≦ u ≦ 5 π / 12 \pi/12\leq u\leq5\pi/12 π /12 ≦ u ≦ 5 π /12 d x = 2 cos u d u dx=2\cos u\,du d x = 2 cos u d u 4 − x 2 = 2 cos u \sqrt{4-x^2}=2\cos u 4 − x 2 = 2 cos u ∫ p q 4 − x 2 d x = ∫ π / 12 5 π / 12 4 cos 2 u d u = [ 2 u + sin 2 u ] π / 12 5 π / 12 = 2 π 3 . \displaystyle
\begin{aligned}
\int_p^q\sqrt{4-x^2}\,dx
&=\int_{\pi/12}^{5\pi/12}4\cos^2u\,du\\
&=[2u+\sin2u]_{\pi/12}^{5\pi/12}=\frac{2\pi}{3}.
\end{aligned} ∫ p q 4 − x 2 d x = ∫ π /12 5 π /12 4 cos 2 u d u = [ 2 u + sin 2 u ] π /12 5 π /12 = 3 2 π . p q = 1 pq=1 pq = 1 q / p = q 2 = 2 + 3 q/p=q^2=2+\sqrt3 q / p = q 2 = 2 + 3 S = 2 π 3 − ln ( 2 + 3 ) . \displaystyle
S=\frac{2\pi}{3}-\ln(2+\sqrt3). S = 3 2 π − ln ( 2 + 3 ) .
双曲線上の点 A の偏角を α \alpha α A = ( r cos α , r sin α ) A=(r\cos\alpha,r\sin\alpha) A = ( r cos α , r sin α ) x y = 1 xy=1 x y = 1 − 1 x 2 = − y x = − tan α \displaystyle -\frac1{x^2}=-\frac yx=-\tan\alpha − x 2 1 = − x y = − tan α min ( 2 α , π − 2 α ) \displaystyle \min(2\alpha,\pi-2\alpha) min ( 2 α , π − 2 α ) π 6 \displaystyle \frac\pi6 6 π α = π 12 \displaystyle \alpha=\frac\pi{12} α = 12 π 5 π 12 \displaystyle \frac{5\pi}{12} 12 5 π x y = r 2 2 sin 2 α = 1 \displaystyle xy=\frac{r^2}{2}\sin2\alpha=1 x y = 2 r 2 sin 2 α = 1 r 2 = 2 sin 2 α = 4 \displaystyle r^2=\frac2{\sin2\alpha}=4 r 2 = sin 2 α 2 = 4 C 2 C_2 C 2 x 2 + y 2 = 4 x^2+y^2=4 x 2 + y 2 = 4
円と双曲線の交点のx座標は x 2 + 1 x 2 = 4 \displaystyle x^2+\frac1{x^2}=4 x 2 + x 2 1 = 4 x > 0 x>0 x > 0 x 4 − 4 x 2 + 1 = 0 x^4-4x^2+1=0 x 4 − 4 x 2 + 1 = 0 x 2 = 2 ± 3 x^2=2\pm\sqrt3 x 2 = 2 ± 3 p = 2 − 3 , q = 2 + 3 p=\sqrt{2-\sqrt3},\quad q=\sqrt{2+\sqrt3} p = 2 − 3 , q = 2 + 3 0 < p < q , p q = 1 0<p<q,\quad pq=1 0 < p < q , pq = 1 p ≦ x ≦ q p\le x\le q p ≦ x ≦ q F ( x ) = 4 − x 2 − 1 x \displaystyle F(x)=\sqrt{4-x^2}-\frac1x F ( x ) = 4 − x 2 − x 1 F ( 1 ) = 3 − 1 > 0 F(1)=\sqrt3-1>0 F ( 1 ) = 3 − 1 > 0
面積は S = ∫ p q 4 − x 2 d x − ln q p \displaystyle S=\int_p^q\sqrt{4-x^2}\,dx-\ln\frac qp S = ∫ p q 4 − x 2 d x − ln p q p 2 = sin π 12 \displaystyle \frac p2=\sin\frac\pi{12} 2 p = sin 12 π q 2 = sin 5 π 12 \displaystyle \frac q2=\sin\frac{5\pi}{12} 2 q = sin 12 5 π x = 2 sin u x=2\sin u x = 2 sin u π 12 ≦ u ≦ 5 π 12 \displaystyle \frac\pi{12}\le u\le\frac{5\pi}{12} 12 π ≦ u ≦ 12 5 π d x = 2 cos u d u dx=2\cos u\,du d x = 2 cos u d u 4 − x 2 = 2 cos u \sqrt{4-x^2}=2\cos u 4 − x 2 = 2 cos u ∫ p q 4 − x 2 d x = ∫ π / 12 5 π / 12 4 cos 2 u d u = [ 2 u + sin 2 u ] π / 12 5 π / 12 = 2 π 3 \displaystyle \int_p^q\sqrt{4-x^2}\,dx=\int_{\pi/12}^{5\pi/12}4\cos^2u\,du=[2u+\sin2u]_{\pi/12}^{5\pi/12}=\frac{2\pi}{3} ∫ p q 4 − x 2 d x = ∫ π /12 5 π /12 4 cos 2 u d u = [ 2 u + sin 2 u ] π /12 5 π /12 = 3 2 π p q = 1 pq=1 pq = 1 q p = q 2 = 2 + 3 \displaystyle \frac qp=q^2=2+\sqrt3 p q = q 2 = 2 + 3 S = 2 π 3 − ln ( 2 + 3 ) \displaystyle S=\frac{2\pi}{3}-\ln(2+\sqrt3) S = 3 2 π − ln ( 2 + 3 )
