Find a point on the curve

Given:

The curve is $y=3 x^{2}-9 x+8$

Differentiating the above w.r.t $x$

$\Rightarrow y=3 x^{2}-9 x+8$

$\Rightarrow \frac{d y}{d x}=2 \times 3 x^{2}-1-9+0$

$\Rightarrow \frac{d y}{d x}=6 x-9 \ldots(1)$

Since, the tangent are equally inclined with axes

i.e, $\theta=\frac{\pi}{4}$ or $\theta=\frac{-\pi}{4}$

$\therefore \frac{d y}{d x}=$ The Slope of the tangent $=\tan \theta$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\tan \left(\frac{\pi}{4}\right)$ or $\tan \left(\frac{-\pi}{4}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1$ or $-1 \ldots(2)$

$\therefore \tan \left(\frac{\pi}{4}\right)=1$

From (1) & (2),we get,

$\Rightarrow 6 x-9=1$ or $6 x-9=-1$

$\Rightarrow 6 x=10$ or $6 x=8$

$\Rightarrow x=\frac{10}{6}$ or $x=\frac{8}{6}$

$\Rightarrow x=\frac{5}{3}$ or $x=\frac{4}{3}$

Substituting $x=\frac{5}{3}$ or $x=\frac{4}{3}$ in $y=3 x^{2}-9 x+8$, we get,

When $x=\frac{5}{3}$

$\Rightarrow y=3\left(\frac{5}{3}\right)^{2}-9\left(\frac{5}{3}\right)+8$

$\Rightarrow y=3\left(\frac{25}{9}\right)-\left(\frac{45}{3}\right)+8$

$\Rightarrow y=\left(\frac{75}{9}\right)-\left(\frac{45}{3}\right)+8$

taking LCM $=9$

$\Rightarrow y=\left(\frac{(75 \times 1)-(45 \times 3)+(8 \times 9)}{9}\right)$

$\Rightarrow y=\left(\frac{75-135+72}{9}\right)$

$\Rightarrow y=\left(\frac{12}{9}\right)$

$\Rightarrow y=\left(\frac{4}{3}\right)$

when $x=\frac{4}{3}$

$\Rightarrow y=3\left(\frac{4}{3}\right)^{2}-9\left(\frac{4}{3}\right)+8$

$\Rightarrow y=3\left(\frac{16}{9}\right)-\left(\frac{36}{3}\right)+8$

$\Rightarrow y=\left(\frac{48}{9}\right)-\left(\frac{36}{3}\right)+8$

taking LCM $=9$

$\Rightarrow y=\left(\frac{(48 \times 1)-(36 \times 3)+(8 \times 9)}{9}\right)$

$\Rightarrow y=\left(\frac{48-108+72}{9}\right)$

$\Rightarrow y=\left(\frac{12}{9}\right)$

$\Rightarrow y=\left(\frac{4}{3}\right)$

Thus, the required point is $\left(\frac{5}{3}, \frac{4}{3}\right) \&\left(\frac{4}{3}, \frac{4}{3}\right)$