# let c be the curve y=x3 .the tangent at A meets the curve again at B .IF GRADIENT AT B IS K TIMES GRADIENT AT A.THEN K=--------

Latika Leekha
8 years ago
Hello student,
Assume that there is a point that lies on the graph and then we will find the equation of the tangent at the that point and henceforth we shall solve this tangent with the given curve to figure out the point where it cuts again and finally we shall find the slope of the curve at that point to find the value of k.
Given curve is y = x3
Tangent at point (say) (1, 1) which lies on the given curve has slope dy/dx = 3x2
So slope at point (1,1) = 3.12 = 3
Equation of tangent at (1,1) is y – 1 = 3(x-1)
So, y = 3x-2.
Solving this and y = x3, we have 3x-2 = x3
or x3 – 3x + 2 = 0
The factors of constant term are ±1, ±2.
x = 1 and x = -2 are the solutions. But, out of these x = 1 has been already included and hence x = -2.
So, y = (-2)3 = -8.
So, the other coordinate would be C(-2, 8).
Gradient at (-2, 8) = 3x2 = 3.4 = 12.
Gradient at C is given to be k times gradient at A.