Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 2 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 2 Maths Textbook Solution.

Answers (1)

ANSWER: Equation of tangent, y-\left(\frac{4}{5}\right)=\left(\frac{13}{16}\right)\left[x-\left(\frac{2 a}{5}\right)\right]

                 Equation  of normal , y-\left(\frac{4}{5}\right)=\left(\frac{-16}{13}\right)\left[x-\left(\frac{2 a}{5}\right)\right]

HINTS:

 Differentiate the given equation   with respect to t  and to get the slopes.

GIVEN:

x=\frac{2 a t^{2}}{1+t^{2}}, y=\frac{2 a t^{2}}{1+t^{2}} \text { at } t=\frac{1}{2}

SOLUTION:

Upon differentiation,

\begin{aligned} &\frac{d x}{d t}=\frac{\left(1+t^{2}\right) y a t-2 a t^{2}(2 t)}{\left(1+t^{2}\right)^{2}} \Rightarrow \frac{d x}{d t}=\frac{4 a t}{\left(1+t^{2}\right)^{2}} \\ &\Rightarrow \frac{d y}{d t}=\frac{\left(1+t^{2}\right) 6 a t^{2}-2 a t^{2}(2 t)}{\left(1+t^{2}\right)^{2}} \Rightarrow \frac{d x}{d t}=\frac{6 a t^{2}+4 a t^{4}}{\left(1+t^{2}\right)^{2}} \\ &\therefore \frac{d y}{d x}=\frac{\left(6 a t^{2}+2 a t^{4}\right)}{4 a t} \end{aligned}

m(\text { tangent }) \text { at } t=\frac{1}{2} \text { is } \frac{13}{16}

The normal is perpendicular to tangent, therefore , m_{1} m_{2}=-1

m(\text { normal }) \text { at } t=\left(\frac{1}{2}\right) \text { is }-\frac{16}{13}

The equation of tangent is given by,

\begin{aligned} &y-y_{1}=m(\text { tangent })\left[x-x_{1}\right] \\ &\Rightarrow y-\left[\frac{4}{5}\right]=\left(\frac{13}{16}\right)\left[x-\left(\frac{2 a}{5}\right)\right] \end{aligned}

The equation of Normal  is given by  ,

\begin{aligned} &y-y_{1}=m(\text { normal })\left[x-x_{1}\right] \\ &\Rightarrow y-\left[\frac{4}{5}\right]=\left(\frac{-16}{13}\right)\left[x-\left(\frac{2 a}{5}\right)\right] \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