The point on the curve y^2=x where tangent makes 45° angle with x-axi…

Let $\left(x_1, y_1\right)$ be the required point on the curve $y^2=x$. Then,

$\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=1$

$\Rightarrow \frac{1}{2 y_1}=1$               $\left[∵ y^2=x \Rightarrow 2 y \frac{d y}{d x}=1 \Rightarrow \frac{d y}{d x}=\frac{1}{2 y}\right]$

$\Rightarrow y_1=\frac{1}{2}$

Since $\left(x_1, y_1\right)$ lies on $y^2=x$.

∴  $x_1=y_1{ }^2 \Rightarrow x_1=\frac{1}{4}$

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