Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 15 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 15 Maths Textbook Solution.

Answers (1)

Answer:

               Point: \left ( 1,0 \right )

Hint:

  1. First we will find the slope of line x-y= 0 then evaluate with the slope of curve
  2. If tangent is perpendicular to the line then  m_{1} \cdot m_{2}=-1

Given:

Given curve  y=x^{2}-3 x+2  where the tangent is perpendicular to the line  x-y= 0

To find:

We have to find the co-ordinate of the point on the given curve.

Solution: 

We have,

               \begin{aligned} & x-y=0 \\ \Rightarrow \quad & y=x \end{aligned}                                                                                                                                                             … (i)

Comparing equation (i) with equation  y=m x+c

 \Rightarrow \quad m_{1}=1

Again, if the line is perpendicular to the tangent then

 \begin{aligned} &\Rightarrow \quad m_{1} \cdot m_{2}=-1 \\ &\Rightarrow \quad 1 \cdot m_{2}=-1 \end{aligned}                                                                                                                                          \left[\because m_{1}=1\right]

\Rightarrow \quad m_{2}=-1

Given curve,

         y=x^{2}-3 x+2                                                                                                                                                  … (ii)

On differentiating both side with respect to x, we get

\Rightarrow \quad \frac{d y}{d x}=2 x-3                                                                                                                        \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

Here slope of tangent is -1

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

Substituting the value of x in equation (ii), we get

          \begin{aligned} &y=1-3+2 \\ &y=0 \end{aligned}

Hence the point is \left ( x,y \right )= \left ( 1,0 \right )

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question