Find a point on the curve y = x2 where the Slope of the tangent is equ

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: The curve is y = x² y = x² Differentiating the above w.r.t x ⇒ \(\frac{dy}{dx}\)= 2x^{2 – 1} ⇒ \(\frac{dy}{dx}\) = 2x ...(1) Also given the Slope of the tangent is equal to the x – coordinate, \(\frac{dy}{dx}\) = x ...(2) From (1) & (2),we get, i.e,2x = x ⇒ x = 0. Substituting this in y = x², we get, y = 0² ⇒ y = 0 Thus, 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:

The curve is y = x²

y = x²

Differentiating the above w.r.t x

⇒ \(\frac{dy}{dx}\)= 2x^{2 – 1}

⇒ \(\frac{dy}{dx}\) = 2x ...(1)

Also given the Slope of the tangent is equal to the x – coordinate,

\(\frac{dy}{dx}\) = x ...(2)

From (1) & (2),we get,

i.e,2x = x

⇒ x = 0.

Substituting this in y = x², we get,

y = 0²

⇒ y = 0

Thus, the required point is (0,0)