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#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 10 Maths Textbook Solution.

Answer:$\text { The required points are }\left(\frac{5}{3}, \frac{4}{3}\right) \text { or }\left(\frac{4}{3}, \frac{4}{3}\right)$

Hint: $\text { The slope of the tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }$

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

Solution: $y=3 x^{2}-9 x+8$

$\text { Differentiating the above with respect to } x$

$\therefore \frac{d}{d x}\left(\mathrm{x}^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0$

$\frac{d y}{d x}=3(2) x^{2-1}-9$

$\frac{d y}{d x}=6 x-9 \quad \rightarrow(1)$

$\frac{d y}{d x}=\text { The slope of tangent }=\tan \theta$

$\text { The tangent is equally inclined with the axis } \theta=\frac{\pi}{4} \text { or } \frac{-\pi}{4}$

$\frac{d y}{d x}=\tan \left(\frac{\pi}{4}\right) \text { or } \tan \left(-\frac{\pi}{4}\right)$

$=1 \text { or }-1 \rightarrow(2)$

From(1)&(2) we get

$6 x-9=1 \text { or } 6 x-9=-1$

$6 x=1+9 \text { or } 6 x=-1+9$

$6 x=10 \quad \text { or } 6 x=8$

$x=\frac{10}{6} \quad \text { or } \quad x=\frac{8}{6}$

$x=\frac{5}{3} \quad \text { or } \quad x=\frac{4}{3}$

Thus the requried points are$\left(\frac{5}{3}, \frac{4}{3}\right) \text { or }\left(\frac{4}{3}, \frac{4}{3}\right)$