Area of the ellipse `x^2/a^2+y^2/b^2=1`is `piab`

The locus of mid-points of a focal chord of the ellipse x^2/a^2+y^2/b^2=1

Find the maximum area of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 which touches the line y=3x+2.

A normal inclined at 45^@ to the axis of the ellipse x^2 /a^2 + y^2 / b^2 = 1 is drawn. It meets the x-axis & the y-axis in P & Q respectively. If C is the centre of the ellipse, show that the area of triangle CPQ is (a^2 - b^2)^2/(2(a^2 +b^2)) sq units

The length ofthe normal (terminated by the major axis) at a point of the ellipse x^2/a^2+y^2/b^2=1 is

The locus of mid point of tangent intercepted between area of ellipse x^2/a^2 +y^2/b^2 = 1 is

A O B is the positive quadrant of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 in which O A=a ,O B=b . Then find the area between the arc A B and the chord A B of the ellipse.

Let d be the perpendicular distance from the centre of the ellipse x^2/a^2+y^2/b^2=1 to the tangent drawn at a point P on the ellipse. If F_1 & F_2 are the two foci of the ellipse, then show the (PF_1-PF_2)^2=4a^2[1-b^2/d^2] .

If the area of the ellipse ((x^2)/(a^2))+((y^2)/(b^2))=1 is 4pi , then find the maximum area of rectangle inscribed in the ellipse.

If the area of the ellipse ((x^2)/(a^2))+((y^2)/(b^2))=1 is 4pi , then find the maximum area of rectangle inscribed in the ellipse.

Given a > b >0,a_1> b_1>0, area of ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 is twice the area of ellipse (x^2)/(a_1^ 2)+(y^2)/(b_1^ 2)=1 . If eccentricity of the ellipse is same then: (A) a=sqrt(2)a_1 (B) a a_1=bb-1 (C) a b=a_1b_1 (D) none of these