# Find the points on the curve

Question:

Find the points on the curve $y=3 x^{2}-9 x+8$ at which the tangents are equally inclined with the axes.

Solution:

Let $\left(x_{1}, y_{1}\right)$ 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 $\pm 45^{\circ}$.

$\therefore$ Slope of the tangent $=\tan (\pm 45)=\pm 1$...(1)

Since, the point lies on the curve.

Hence, $y_{1}=3 x_{1}^{2}-9 x_{1}+8$

Now, $y=3 x^{2}-9 x+8$

$\Rightarrow \frac{d y}{d x}=6 x-9$

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=6 x_{1}-9$ $\ldots(2)$

From eq. (1) and eq. (2), we get

$6 x_{1}-9=\pm 1$

$\Rightarrow 6 x_{1}-9=1$ or $6 x_{1}-9=-1$

$\Rightarrow 6 x_{1}=10$ or $6 x_{1}=8$

$\Rightarrow x_{1}=\frac{10}{6}=\frac{5}{3}$ or $x_{1}=\frac{8}{6}=\frac{4}{3}$

Also,

$y_{1}=3\left(\frac{5}{3}\right)^{2}-9\left(\frac{5}{3}\right)+8$ or $y_{1}=3\left(\frac{4}{3}\right)^{2}-9\left(\frac{4}{3}\right)+8$

$\Rightarrow y_{1}=\frac{25}{3}-\frac{45}{3}+8$ or $y_{1}=\frac{16}{3}-\frac{36}{3}+8$

$\Rightarrow y_{1}=\frac{4}{3}$ or $y_{1}=\frac{4}{3}$

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