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

8965 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Differentiate the following equation: f(x) = 5x^3 + 6x^2 - 12x + 4


Differentiate with respect to x, x^2*e^(tan(x))


Express 9^(3x+)1 in the form 3^y giving y in the form of ax+b where a and b are constants.


Solve 29cosh x – 3cosh 2x = 38 for x, giving answers in terms of natural logarithms


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning