Find where the curve 2x^2 + xy + y^2 = 14 has stationary points

d/dx (xy) = x dy/dx + y 

d/dx (y^2) = 2y dy/dx [This is from the chain rule]

So, d/dx (2x^2 + xy + y^2 = 14) 

=> 4x + x dy/dx + y + 2y dy/dx = 0

set dy/dx = 0 as stationary point has gradient 0

Obtains 4x+y=0

y=-4x

Sub this back into our original equation

14x^2 = 14

x^2 = 1

This is only satisfied by +1 and -1

When x=1 y=-4, when x=-1 y=4

So stationary points are (1,-4) and (-1,4)

MH
Answered by Matthew H. Maths tutor

7916 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Use the chain rule to show that, if y = sec(x), then dy/dx = sec(x)tan(x).


Find the exact solutions, in their simplest form, to the equations : a) 2ln(2x + 1)-4=0 b)7^(x)e^(4x)=e^5


By using partial fractions, integrate the function: f(x) = (4-2x)/(2x+1)(x+1)(x+3)


Find a solution for the differential equation dy/dx=exp(-y)*sin2x which passes through the origin.


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