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)

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence, } y_1 = 3 {x_1}^2 - 9 x_1 + 8 \]

\[\text { Now, } y = 3 x^2 - 9x + 8\]

\[ \Rightarrow \frac{dy}{dx} = 6x - 9\]

\[\text { Slope of the tangent at}\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =6 x_1 -9 .......(2)\]

\[\text { From eq. (1) and eq. (2), we get }\]

\[6 x_1 - 9 = \pm 1\]

\[ \Rightarrow 6 x_1 - 9 = 1 \text { or }6 x_1 - 9 = - 1\]

\[ \Rightarrow 6 x_1 = 10 \text { or }6 x_1 = 8\]

\[ \Rightarrow x_1 = \frac{10}{6} = \frac{5}{3} \text { or }x_1 = \frac{8}{6} = \frac{4}{3}\]

\[\text { Also,} \]

\[ y_1 = 3 \left( \frac{5}{3} \right)^2 - 9\left( \frac{5}{3} \right) + 8 \text { or } y_1 = 3 \left( \frac{4}{3} \right)^2 - 9\left( \frac{4}{3} \right) + 8\]

\[ \Rightarrow y_1 = \frac{25}{3} - \frac{45}{3} + 8 \text { or } y_1 = \frac{16}{3} - \frac{36}{3} + 8\]

\[ \Rightarrow y_1 = \frac{4}{3} \text { or } y_1 = \frac{4}{3}\]

\[\text { Thus, the required points are }\left( \frac{5}{3}, \frac{4}{3} \right)\text { and }\left( \frac{4}{3}, \frac{4}{3} \right).\]