The points on the ellipse `(x^(2))/(25)+(y^(2))/(9)=1` Whose eccentric angles differ by a right angle are

If the eccentric angles of two points P and Q on the ellipse (x^(2))/(28)+(y^(2))/(7)=1 whose centre is C differ by a right angle then the area of Delta CPQ is

P and Q are two points on the ellipse (x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1 whose eccentric angles are differ by 90^(@) , then

The area of the triangle formed by three points on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 whose eccentric angles are alpha,beta and gamma is

Area of the triangle formed by the tangents at the point on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 whose eccentric angles are alpha,beta,gamma is

The locus of the point of intersection of tangents to the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 at the points whose eccentric angles differ by pi//2 , is

Find the condition so that the line px+qy=r intersects the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 in points whose eccentric angles differ by (pi)/(4) .

The ecentric angle of the points on the ellipse (x^(2))/(6)+(y^(2))/(2)=1 whose distane from its centre is 2 is/are.

A point P on the ellipse (x^2)/(25)+(y^2)/(9)=1 has the eccentric angle (pi)/(8) . The sum of the distances of P from the two foci is d. Then (d)/(2) is equal to:

The line 3x+5y-15sqrt(2) is a tangent to the ellipse (x^(2))/(25)+(y^(2))/(9)=1 at a point whose eccentric angle is

Find the sum of squares of two distances of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 whose extremities have eccentric angles differ by (pi)/(2) and sow that is a constant.