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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 17 Maths Textbook Solution.

Answers (1)

Answer:

                (b) Is the correct option

Hint:

        Substituting x=1in x=3t^{2}-1

Given:

            x=3t^{2}+1,y=t^{3}-1

Solution:

x=3t^{2}+1,                            (1)

Differentiate w.r.t t

\begin{aligned} &\frac{d x}{d t}=6 t\\ &\text { And, } y=t^{3}-1\\ &\text { Differentiating w.r.t t }\\ &\frac{d y}{d x}=3 t^{2} \end{aligned}

\begin{aligned} &\text { Substituting } x=1 \text { in (1) }\\ &1=3 t^{2}+1\\ &3 t^{2}=0\\ &t=0\\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 t^{2}}{6 t}=\frac{t}{2}\\ &\left[\frac{d y}{d x}\right]_{t=0}=0\\ &\therefore \text { slope }=0 \end{aligned}

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