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

9218 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

a) Find the indefinite integral of sec^2(3x) with respect to x. b) Using integration by parts, or otherwise, find the indefinite integral of x*sec^2(3x) with respect to x.


Evaluate the indefinite integral: ∫ (e^x)sin(x) dx


(The question is too long so it's marked at the top of the answer space, sorry for any inconveniences)


The curve y = 2x^3 - ax^2 + 8x + 2 passes through the point B where x=4. Given that B is a stationary point of the curve, find the value of the constant a.


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