Let E be an ellipse x^2 a^2 + y^2 b^2 = 1 ( a > b > 0 ) with origin O . The area of the director circle of E is 34 . If OP and OQ are any two mutually perpendicular semi-diameters of E such that 1 OP^2 + 1 OQ^2 = 34 225 , then the value of 5L , where L is the length of the latus rectum of E , is

Step-by-step solution

The area of the director circle of the ellipse x^2 a^2 + y^2 b^2 = 1 is given by (a^2 + b^2) . Thus, (a^2 + b^2) = 34 a^2 + b^2 = 34 . For any two mutually perpendicular semi-diameters OP and OQ of the ellipse, we have the standard property: 1 OP^2 + 1 OQ^2 = 1 a^2 + 1 b^2 Given that this sum is 34 225 , we get: a^2 + b^2 a^2 b^2 = 34 225 Substituting a^2 + b^2 = 34 : 34 a^2 b^2 = 34 225 a^2 b^2 = 225 Since a^2 + b^2 = 34 and a^2 b^2 = 225 , a^2 and b^2 are the roots of the quadratic equation t^2 - 34t + 225 = 0 .