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

7803 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Where do the graphs of y=3x-2 and y=x^2+4x-8 meet?


Find the coordinates of the stationary points y=x^4-8x^2+3


A curve has equation y = (x-1)e^(-3x). The curve has a stationary point M. Show that the x-coordinate of M is 4/3.


f(x) = (sin(x))^3. What is f'(x)


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