Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

18.    At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answers (1)

best_answer

Slope m of  line joining (x,y) and (-4,-3) is \frac{y+3}{x+4}

According to the question,

\\ \frac{dy}{dx} = 2(\frac{y+3}{x+4}) \\ \implies \int \frac{dy}{y+3} = 2\int \frac{dx}{x+4} \\ \implies \log (y+3) = 2\log (x+4) + \log k \\ \implies (y+3) = k(x+4)^2

Now, Since the curve passes through (-2,1)

x = -2 , y =1

\\ \implies (1+3) = k(-2+4)^2 \\ \implies k =1

Putting the value of k, we get

\\ \implies y+3 = (x+4)^2

Posted by

HARSH KANKARIA

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

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question