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 9 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 9 Maths Textbook Solution.

Answers (1)

ANSWER: Equation of tangent: y+\left ( \frac{71}{4} \right )=3\left \{ x+\left ( \frac{1}{2} \right ) \right \}

HINTS:

 Differentiate  with respect to x   to get its slope

GIVEN:

y=x^{2}+4x-16 Which is parallel to 3x-y+1=0

SOLUTION:

Upon differentiation

\frac{dy}{dx}=2x+4\Rightarrow m\left ( tangent \right )=2x+4

The equation of tangent is given by y-y_{1}=m\left ( tangent \right )\left [ x-x_{1} \right ]

Comparing the slopes of a tangent with the given equation

2x+y=3\Rightarrow x=-\frac{1}{2}

Substitutes the value of x  in the curve to find y

y=\left ( \frac{1}{4} \right )-2-16=-\frac{71}{4}

So,the equation of tangent is parallel to the given line is

y+\left ( \frac{71}{4} \right )=3\left \{ x+\left ( \frac{1}{2} \right ) \right \}

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question