Question

Find the points on the curve y = 3x² − 9x + 8 at which the tangents are equally inclined with the axes.

Answer 1

Let (x₁, y₁) be the required point.
It is given that the tangent at this point is equally inclined to the axes. It means that the angle made by the tangent with the x-axis is $\pm$45$°$.
∴ Slope of the tangent = tan ($\pm$45) = $\pm$ 1 ...(1)

$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1=3{x_1}^2-9x_1+8 \\ \mathrm{Now},y=3x^2-9x+8 \\ \Rightarrow \frac{dy}{dx}=6x-9 \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=6}{\mathit{\text{x}}}_{\mathit{1}}\text{-9 ...(2)} \\ \text{From eq. (1) and eq. (2), we get} \\ 6x_1-9=\pm 1 \\ \Rightarrow 6x_1-9=1\text{or}6x_1-9=-1 \\ \Rightarrow 6x_1=10\text{or}6x_1=8 \\ \Rightarrow x_1=\frac{10}{6}=\frac{5}{3}\text{or}{x}_1=\frac{8}{6}=\frac{4}{3} \\ \mathrm{Also}, \\ {y}_1=3{\left(\frac{5}{3}\right)}^2-9\left(\frac{5}{3}\right)+8\text{or}{y}_1=3{\left(\frac{4}{3}\right)}^2-9\left(\frac{4}{3}\right)+8 \\ \Rightarrow y_1=\frac{25}{3}-\frac{45}{3}+8\text{or}{y}_1=\frac{16}{3}-\frac{36}{3}+8 \\ \Rightarrow y_1=\frac{4}{3}\text{or}{y}_1=\frac{4}{3} \\ \mathrm{Thus},\mathrm{the}\mathrm{required}\text{points are}\left(\frac{5}{3},\frac{4}{3}\right)\text{and}\left(\frac{4}{3},\frac{4}{3}\right)\text{.} \end{gathered}$$