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Problem If x a cos α + y b sin α = 1 \frac{x}{a}\cos\alpha+\frac{y}{b}\sin\alpha=1 a x cos α + b y sin α = 1 x a cos β + y b sin β = 1 \frac{x}{a}\cos\beta+\frac{y}{b}\sin\beta=1 a x cos β + b y sin β = 1 cos α cos β a 2 + sin α sin β b 2 = 0 \frac{\cos\alpha\cos\beta}{a^2}+\frac{\sin\alpha\sin\beta}{b^2}=0 a 2 c o s α c o s β + b 2 s i n α s i n β = 0 tan α tan β = b 2 ( x 2 + a 2 ) a 2 ( y 2 + b 2 ) \tan\alpha\tan\beta=\frac{b^2(x^2+a^2)}{a^2(y^2+b^2)} tan α tan β = a 2 ( y 2 + b 2 ) b 2 ( x 2 + a 2 ) x 2 + y 2 = a 2 + b 2 x^2+y^2=a^2+b^2 x 2 + y 2 = a 2 + b 2 tan α tan β = a 2 b 2 \tan\alpha\tan\beta=\frac{a^2}{b^2} tan α tan β = b 2 a 2 Browse the problem library