SOLUTION: Find the eccentric angle in the first quadrant of a point on the ellipse {{{ x^2/10 + y^2/8 =1 }}} which is at the distance of 3 units from the center of ellipse?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the eccentric angle in the first quadrant of a point on the ellipse {{{ x^2/10 + y^2/8 =1 }}} which is at the distance of 3 units from the center of ellipse?      Log On


   

Question 1184208: Find the eccentric angle in the first quadrant of a point on the ellipse +x%5E2%2F10+%2B+y%5E2%2F8+=1+ which is at the distance of 3 units from the center of ellipse?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given:
the ellipse +x%5E2%2F10+%2B+y%5E2%2F8+=1+
a point is at the distance of 3 units from the center of ellipse

let a point P(a%2Acos+%28theta%29 ,b%2Asin%28theta%29 )
in
a=sqrt%2810%29
b=sqrt%288%29
P(sqrt%2810%29cos%28theta%29 ,sqrt%288%29sin%28theta%29 )

its distance from (0,0) is 3
10cos%5E2+%28theta%29+%2B8sin%5E2%28theta+%29=9
18sin%5E2%28theta+%29=9....divide by 9
2cos%5E2%28theta%29+=1
cos%5E2%28theta%29+=1%2F2
cos%28theta%29+=1%2Fsqrt%282%29
theta+=pi%2F4