Question

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

Answer 1

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},{y_1}^2=x_1 \\ \mathrm{Now},y^2=x \\ \Rightarrow 2y\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2y} \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{1}{2y_1} \\ \text{Given:} \\ \frac{1}{2y_1}=1 \\ \Rightarrow 2y_1=1 \\ \Rightarrow y_1=\frac{1}{2} \\ \mathrm{Now}, \\ {x}_1={y_1}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4} \\ \therefore \left(x_1,y_1\right)=\left(\frac{1}{4},\frac{1}{2}\right) \end{gathered}$$