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

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

Answers (1)

Answer:

              \left ( b \right )\frac{6}{7}

Hint:

        Use differentiation

Given:

            x=t^{2}+3t-8

            y=2t^{2}+2t-5

Solution:

Given curve are x=t^{2}+3t-8            (1)

And y=2t^{2}+2t-5                           (2)

At \left ( 2,-1 \right )

From (1) t^{2}+3t-10=0

\Rightarrow t=2 \; or\: t=-5

From (2) 2t^{2}-2t-4=0

\Rightarrow t^{2}-t-2=0

\Rightarrow t=2 \: \: or \: \: t=-1

From both the solution, we get t=2

Differentiating both the equation w.r.t t, we get

\frac{dx}{dt}=2t+3                                                    (3)

\frac{dy}{dt}=4t-2                                                        (4)

Now,\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

From (3) and (4) we get

\frac{4t-2}{2t+3}

\therefore \frac{dy}{dx}=\frac{4t-2}{2t+3}     Is the slope of the tangent to the given curve.

\left|\frac{d y}{d x}\right|_{(2,-1)}=\left|\frac{4 t-2}{2 t+3}\right|_{t-2}=\frac{8-2}{4+3}=\frac{6}{7}         Is the slope of the tangent to the given curve at (2,-1)

 

 

                

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