Question

The point on the curve y² = x, the tangent at which makes an angle of 45^o with x-axis is ____________________.

Answer 1

Let (h, k) be the point on the curve y² = x, the tangent at which makes an angle of 45º with x-axis.

y² = x

Differentiating both sides with respect to x, we get

$2y\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}$

∴ Slope of tangent at (h, k) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=\frac{1}{2k}$ .....(1)

It is given that, the tangent makes an angle of 45º with x-axis.

∴ Slope of the tangent = tan45º = 1 .....(2)

From (1) and (2), we have

$\frac{1}{2k}=1$

$\Rightarrow k=\frac{1}{2}$

Now, (h, k) lies on the curve.

∴ k² = h .....(3)

Putting $k=\frac{1}{2}$ in (3), we get

$h={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

So, the coordinates of the required point are $\left(\frac{1}{4},\frac{1}{2}\right)$.

Thus, the point on the curve y² = x, the tangent at which makes an angle of 45º with x-axis is $\left(\frac{1}{4},\frac{1}{2}\right)$.


The point on the curve y² = x, the tangent at which makes an angle of 45^o with x-axis is $\underline{\left(\frac{1}{4},\frac{1}{2}\right)}$.