Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 6 Maths Textbook Solution.

Answer: $\text { The required point is }(2,4)$

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

Given:$\text { The curve } y^{2}=2 x^{3} \text { and the slope of tangent is } 3$

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

\begin{aligned} &\text { Differentiating the above with respect to } x\\ &\therefore \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\\ &2 y^{2-1} \frac{d y}{d x}=2 \times(3 x)^{3-1} \end{aligned}

\begin{aligned} &2 y \frac{d y}{d x}=2 \times 3 x^{2} \\ &y \frac{d y}{d x}=3 x^{2} \\ &\frac{d y}{d x}=\frac{3 x^{2}}{y} \end{aligned}

Since the slope of tangent is 3

\begin{aligned} &\frac{3 x^{2}}{y}=3 \\ &\frac{x^{2}}{y}=1 \\ &x^{2}=y \end{aligned}

$\text { Substituting } x^{2}=y \text { in } y^{2}=2 x^{3}$

$\left(x^{2}\right)^{2}=2 x^{3}$

$x^{4}-2 x^{3}=0$

$x^{3}(x-2)=0$

$x^{3}=0 \quad \text { or } \quad x-2=0$

$x=0 \quad \text { or } \quad x=2$

$\text { If } x=0$

$\frac{d y}{d x}=\frac{3(0)^{2}}{y}$

$\frac{d y}{d x}=0 \text { which is not possible }$

$\text { So we take } x=2 \text { and substitute it in } y^{2}=2 x^{3}$

\begin{aligned} &y^{2}=2(2)^{3} \\ &y^{2}=2 \times 8 \\ &y^{2}=16 \end{aligned}

$y=4 \ldots . .\left\{\text { As } y=x^{2},-4 \text { as been discarded }\right\}$

$\text { So,the required point is }(2,4)$