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

8537 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Express (9x^2 + 43x + 8)/(3+x)(1-x)(2x+1) in partial fractions.


Find dy/dx of the equation (x^3)*(y)+7x = y^3 + (2x)^2 +1 at point (1,1)


How do i remember the difference between differentiation and integration?


theta = arctan(5x/2). Using implicit differentiation, find d theta/dx.


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:

© 2025 by IXL Learning