Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 15 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 15 Maths Textbook Solution.

Answers (1)

ANSWER: Equation of tangent, x-y-1=0

                 Equation  of normal , x+y-3=0

 

HINTS:

 Differentiating the given curve with respect to x and find its slope .

GIVEN:

x^{2}=4 y \text { at }(2,1)

SOLUTION:

\begin{aligned} &2 x=4 \frac{d y}{d x} \\ &\Rightarrow \frac{d y}{d x}=\frac{x}{2} \end{aligned}

Slope of tangent,

m=\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{2}{2}=1

Equation of tangent is

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-1=-1(-x+2) \\ &\Rightarrow y-1=x-2 \\ &\therefore x-y-1=0 \end{aligned}

Equation of Normal  is ,

\begin{aligned} &y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-1=1(-x+2) \\ &\Rightarrow y-1=-x+2 \\ &\therefore x+y-3=0 \end{aligned}

Posted by

infoexpert21

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