The length of latusrectum of the parabola `13[(x-3)^(2)+(y-4)^(2)]=(2x-3y+5)^(2)` is
A) `(2)/(sqrt(13))` B) `(4)/(sqrt(13))` C) `(1)/(sqrt(13))` D) `(5)/(sqrt(13))`

The length of the tangent from the point (4, 5) to the circle x^2 + y^2 + 2x - 6y = 6 is : (A) sqrt(13) (B) sqrt(38) (C) 2sqrt(2) (D) 2sqrt(13)

The length of the chord of the parabola y^(2)=x which is bisected at the point (2,1) is (a) 2sqrt(3)( b) 4sqrt(3)(c)3sqrt(2) (d) 2sqrt(5)

sin[(1)/(2)cot^(-1)((2)/(3))]= sqrt((sqrt(13)-2)/(2sqrt(13))) (2+sqrt(13))/(2sqrt(13)) sqrt((2-sqrt(13))/(2sqrt(13))) (2-sqrt(13))/(2sqrt(13))

(13+2sqrt(5))^(2)=?sqrt(5)+189

Simplify (sqrt(13)-sqrt(11))/(sqrt(13)+sqrt(11)) + (sqrt(13)+sqrt(11))/(sqrt(13)-sqrt(11))

Two circles, each having radius 4, have a common tangent given by 3x+2y-6=0 at (2, 0) . Then their centres are : (A) (2+ 5/sqrt(13), 8/sqrt(13)), (2- 5/sqrt(13) , (-8)/sqrt(13)) (B) (2+ 12/sqrt(13), 8/sqrt(13)) , (2- 12/sqrt(13), (-8)/sqrt(13)) (C) (2, 3), (4, 5) (D) none of these

If y=cos^(-1){(2x-3sqrt(1-x^(2)))/(sqrt(13))}, find (dy)/(dx)

-39x^(4) by sqrt(13)x^(2)

Domain of f(x) = sqrt(2x-1) + sqrt(13) cos^(-1) ((2x-1)/2) is

y=sin^(-1)((5x+12sqrt(1-x^(2)))/(13))