The point P on the ellipse `4x^2+9y^2=36` is such that the are of the triangle` P F_1F_2=sqrt(10)` where `F_1,f_2` are foci. Then P has the coordinates `(3/(sqrt(2)),sqrt(2))` (b) `(3/2,2)` `(-3/2,-2)` (d) `(-3/(sqrt(2))-sqrt(2))`

The point P on the ellipse 4x^2+9y^2=36 is such that the are of the P F_1F_2=sqrt(10) where F_1,f_2 are foci. Then P has the coordinates (a) (3/(sqrt(2)),sqrt(2)) (b) (3/2,2) (-3/2,-2) (d) (-3/(sqrt(2))-sqrt(2))

The value of sqrt(3-2\ sqrt(2)) is (a) sqrt(2)-1 (b) sqrt(2)+1 (c) sqrt(3)-sqrt(2) (d) sqrt(3)+\ sqrt(2)

P is any variable point on the ellipse 4x^(2) + 9y^(2) = 36 and F_(1), F_(2) are its foci. Maxium area of trianglePF_(1)F_(2) ( e is eccentricity of ellipse )

1/(sqrt(9)-\ sqrt(8)) is equal to: (a) 3+2sqrt(2) (b) 1/(3+2sqrt(2)) (c) 3-2sqrt(2) (d) 3/2-\ sqrt(2)

(sqrt(3)-sqrt(2))/(sqrt(3)+sqrt(2))-(sqrt(3)+sqrt(2))/(sqrt(3)-sqrt(2))+(1)/(sqrt(2)+1)-(1)/(sqrt(2)-1)

If P and Q are two points on the circle x^2 + y^2 - 4x + 6y-3=0 which are farthest and nearest respectively from the point (7, 2) then. (A) P-=(2-2sqrt(2), -3-2sqrt(2)) (B) Q-= (2+2sqrt(2), -3 + 2 sqrt(2)) (C) P-= (2+2sqrt(2), -3 + 2 sqrt(2)) (D) Q-= (2-2sqrt(2), -3 + 2 sqrt(2))

The eccentricity of the ellipse 4x^2+9y^2+8x+36 y+4=0 is a. 5/6 b. 3/5 c. (sqrt(2))/3 d. (sqrt(5))/3

Let P be a variable point on the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 with foci F_1" and "F_2 . If A is the area of the trianglePF_1F_2 , then the maximum value of A is

Let P be a variable point on the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 with foci F_1" and "F_2 . If A is the area of the trianglePF_1F_2 , then the maximum value of A is

f(x)=(x-1)|(x-2)(x-3)|dot Then f decreases in (a) (2-1/(sqrt(3)),2) (b) (2,2+1/(sqrt(3))) (c) (2+1/(sqrt(3)),4) (d) (3,oo)