A curve has the equation 2x^2 + xy - y^2 +18 = 0. (1) Find the coordinates of the points where the tangent to the curve is parallel to the x-axis.

Firstly notice that the coordinates of the points where the tangent to the curve is parallel to the x-axis are precisely the points where the rate of change of y with respect to x is not changing. That is, when dy/dx = 0. 

So, differentiating euqation (1) with respect to x gives:

4x + y + xdy/dx - 2ydy/dx = 0, where underline denotes the use of implicit differentiation.

Now, dy/dx = 0 gives us:

4x + y + 0, i.e. y = -4x. 

Substituting this back into euqation (1) and solving the quatratic for x gives x = +/- 1. Thus y = -/+ 4.

Hence, coordinates of the points where the tangent to the curve is parallel to the x-axis are (1,-4), (-1,4).

OL
Answered by Oliver L. Maths tutor

7921 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How would I prepare for my Maths exams so that I get the best grade possible?


How would you differentiate 3x^4 - 2x^2 + 9x - 1


Calculate the integral of e^x*sin x


What is the equation of a curve with gradient 4x^3 -7x + 3/2 which passes through the point (2,9)?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences