Tangents to ellipse

Find the magnitude of the angle at which the ellipse x2 + 5 y2 = 5 is visible from the point P[5, 1].

Correct answer:

A =  26.5651 °

Step-by-step explanation:

x2 + 5 y2 = 5 P0=5;P1=1  x2/5 + (y/1)2 = 1 a=5=52.2361 b=1 x0=0 y0=0  p1,p2: y=kx+q b2  q2 + a2   k2 = 0  1q2+5 k2 = 0 P1 = k P0+q  1(P1k P0)2+5 k2=0  1(1k 5)2+5 k2=0 20k2+10k=0 20k210k=0 20=225 10=25 0 GCD(20,10,0)=25=10  2k2k=0  a=2;b=1;c=0 D=b24ac=12420=1 D>0  k1,2=2ab±D=41±1 k1,2=41±1 k1,2=0.25±0.25 k1=0.5 k2=0  q1=P1k1 P0=10.5 5=23=1.5 q2=P1k2 P0=10 5=1  φ1=π180°arctan(k1)=π180°arctan0.526.5651  φ2=π180°arctan(k2)=π180°arctan0=0   A=φ1φ2=26.56510=26.5651=26°3354"

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