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Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise15.1 Question 13 maths Textbook Solution.

Answers (1)

Answer: $\text { The required point is }(1,7)$

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

Given: $\text { The curve is } y=3 x^{2}+4$

Solution: $y=3 x^{2}+4$

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

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

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

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

$\text { The slope } \mathrm{m}_{1}=\frac{d y}{d x}=6 x$

$\text { Now the given slope is } \mathrm{m}_{2}=\frac{-1}{6}$

$\text { We have tangent is perpendicular to tangent whose slope is } \frac{-1}{6}$

$m_{1} \times m_{2}=-1$

$6 x \times \frac{-1}{6}=-1$

$x=1$

From(1)

$y=7$

Thue the required points is (1,7)

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