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

8964 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What is the differential of (14x^3-3x^2)^3


The mass of a substance is increasing exponentially. Initially its mass is 37.5g, 5 months later its mass is 52g. What is its mass 9 months after the initial value to 2 d.p?


Given the equation 0=5x^2+3xy-y^3 find the value of dy/dx at the point (-2,2)


Integrate ((7e^(x/2))/4) with respect to x within the bounds of x=0 and x=2. (Basic introduction to definite integration)


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