Question

equation of tangent drawn to curve y=x square at x=1

Answer 1

(2y-4)(dy/dx)-6x^2=0
=>dy/dx=3x^2/(y-2)--------(1)

(1) is the slope of the tangent to the curve at any point (x,y) on the curve

Since tangents are drawn from (1,2)
Slope of the tangent from (1,2) to (x,y) will be
y-2/x-1 = 3x^2/y-2
=> (y-2)^2 =3x^2(x-1)------(2)
Point (x,y) lies on the curve hence
(y-2)^2=2x^3-4-----(3)
From (2) & (3)
x=-1,2
We neglect -1 since there are no real values of y for x=-1
Now you know the coordinates of the point where tangents are drawn from (1,2) and those are (2,2+√3) & (2,2-√3)