Find a point on the curve y = x2 where the slope of the tangent is eq

1.	Find a point on the curve y = x2 where the slope of the tangent is equal to the x – coordinate of the point.
Answer» Given as the curve is y = x² y = x² Differentiate the above with respect to x (dy/dx)_{= 2x^{2 - 1}} (dy/dx)_{= 2x ...(1)} The slope of tangent is equal to the x-coordinate, (dy/dx)_{= x ...(2)} From the equation (1) & (2), 2x = x ⇒ x = 0. Substitute in y = x², y = 0² ⇒ y = 0 Hence, the required point is (0, 0)

Find a point on the curve y = x2 where the slope of the tangent is equal to the x – coordinate of the point.

Answer»

Given as the curve is y = x²

y = x²

Differentiate the above with respect to x

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

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

The slope of tangent is equal to the x-coordinate,

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

From the equation (1) & (2),

2x = x

⇒ x = 0.

Substitute in y = x²,

y = 0²

⇒ y = 0

Hence, the required point is (0, 0)