Question

At what point of the curve y = x² does the tangent make an angle of 45^° with the x–axis?

Answer 1

Given as the curve is y = x²

Differentiate the above with respect to x

⇒ y = x²

(dy/dx)_{= 2x^{2 - 1}}  

(dy/dx)_{= 2x ...(1)}

So, (dy/dx)_{= The slope of tangent = tan θ}

The tangent make an angle of 45° with x-axis

(dy/dx)_{= tan(45°) = 1 ...(2)}

Because the tan(45°) = 1

From the equation (1) & (2)

2x = 1

x = 1/2

Substitute x = 1/2 in y = x²

y = (1/2)²

y = 1/4 

Hence, the required point is (1/2,1/4)