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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 16 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 16 Maths Textbook Solution.

Answers (1)

Answer: The requried points are (3,2)

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 { Equation of the curve is } y=x^{2}-4 x+5 \quad \rightarrow(1)

\text { Equation of the line is } 2 y+x=7 \quad \rightarrow(2)

Solution:\text { Differentiating eqn(1) 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

\frac{d y}{d x}=2 x^{2-1}-4

\frac{d y}{d x}=2 x-4

\text { Slope } \mathrm{m}_{1}=\frac{d y}{d x}=2 x-4 \quad \rightarrow(3)

\text { Differentiating eqn }(2) \text { with respect to } x

2 \frac{d y}{d x}+1=0

2 \frac{d y}{d x}=-1

\frac{d y}{d x}=\frac{-1}{2}

\text { Slope } \mathrm{m}_{2}=\frac{d y}{d x}=\frac{-1}{2}

\text { We have given that slope(1) and slope (2) are perpendicular to each other }

Hence,

\begin{aligned} &m_{1} \times m_{2}=-1 \\ &(2 x-4) \times\left(\frac{-1}{2}\right)=-1 \\ &(2 x)\left(\frac{-1}{2}\right)-(4)\left(\frac{-1}{2}\right)=-1 \end{aligned}

\begin{aligned} &-x+2=-1 \\ &x=2+1 \\ &x=3 \end{aligned}

\text { When put } x=3 \text { in eqn (1) } y=x^{2}-4 x+5

\begin{aligned} &y=(3)^{2}-4(3)+5 \\ &y=9-12+5 \\ &y=2 \end{aligned}

Thus the required points are (3,2)

 

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