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  • Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 15 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 15 maths Textbook Solution.

Answers (1)

Answer: The requried point are (0,0) \&(2 a,-2 a)

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

Given:The Curve is 2 a^{2} y=x^{3}-3 a x^{2} \rightarrow(1)

Solution:2 a^{2} y=x^{3}-3 a x^{2}

\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

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

2 a^{2} \frac{d y}{d x}=3 x^{2}-6 a x

\frac{d y}{d x}=\frac{1}{2 a^{2}}\left[3 x^{2}-6 a x\right]

\text { Slope } \mathrm{m}_{1}=\frac{d y}{d x}=\frac{1}{2 a^{2}}\left[3 x^{2}-6 a x\right] \rightarrow(2)

Also

\text { Slope } \mathrm{m}_{2}=\frac{d y}{d x}=\tan \theta

\tan 0^{\circ}=0

\text { [slope is parallel to } x \text { -axis] }

m_{1}=m_{2}

\frac{1}{2 a^{2}}\left[3 x^{2}-6 a x\right]=0

3 x^{2}-6 a x=0

\begin{aligned} &3 x(x-2 a)=0 \\ &3 x=0 \text { or } x-2 a=0 \\ &x=0 \text { or } x=2 a \end{aligned}

\text { When put } x=0 \text { in eqn(1) } 2 a^{2} y=x^{3}-3 a x^{2}

\begin{aligned} &2 a^{2} y=(0)^{3}-3 a(0)^{2} \\ &2 a^{2} y=0 \\ &y=0 \end{aligned}

\text { When put } x=2 a \text { in eqn (1) } 2 a^{2} y=x^{3}-3 a x^{2}

\begin{aligned} &2 a^{2} y=(2 a)^{3}-3 a(2 a)^{2} \\ &2 a^{2} y=8 a^{3}-3 a\left(4 a^{2}\right) \\ &2 a^{2} y=8 a^{3}-12 a^{3} \\ &2 a^{2} y=-4 a^{3} \\ &y=-2 a \end{aligned}

Thus the requried points are (0,0) \&(2 a,-2 a)

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