At what point of the curve y = x2 does the tangent make an angle of 45

1.	At what point of the curve y = x2 does the tangent make an angle of 45° with the x–axis?
Answer» Given: The curve is y = x² Differentiating the above w.r.t x ⇒ y = x² ⇒ \(\frac{dy}{dx}\) = 2x^{2 – 1} ⇒ \(\frac{dy}{dx}\) = 2x ...(1) \(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ Since, the tangent make an angle of 45^o with x – axis i.e, ⇒ \(\frac{dy}{dx}\) = tan(45°) = 1 ...(2) \(\therefore\) tan(45°) = 1 From (1) & (2), we get ⇒ 2x = 1 ⇒ x = \(\frac{1}{2}\) Substituting x = \(\frac{1}{2}\) in y = x², we get, ⇒ y = ( \(\frac{1}{2}\) )² ⇒ y = \(\frac{1}{4}\) Thus, the required point is ( \(\frac{1}{2}\), \(\frac{1}{4}\))

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

Answer»

Given:

The curve is y = x²

Differentiating the above w.r.t x

⇒ y = x²

⇒ \(\frac{dy}{dx}\) = 2x^{2 – 1}

⇒ \(\frac{dy}{dx}\) = 2x ...(1)

\(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ

Since, the tangent make an angle of 45^o with x – axis

i.e,

⇒ \(\frac{dy}{dx}\) = tan(45°) = 1 ...(2)

\(\therefore\) tan(45°) = 1

From (1) & (2), we get

⇒ 2x = 1

⇒ x = \(\frac{1}{2}\)

Substituting x = \(\frac{1}{2}\) in y = x², we get,

⇒ y = ( \(\frac{1}{2}\) )²

⇒ y = \(\frac{1}{4}\)

Thus, the required point is ( \(\frac{1}{2}\), \(\frac{1}{4}\))