A curve is described by the equation (x^2)+4xy+(y^2)+27=0. The tangent to the point P, which lies on the curve, is parallel to the x-axis. Given the x-co-ordinate of P is negative, find the co-ordinates of P.

Firstly, implicitly differentiate the function to find dy/dx in terms of y and x: dy=dx = (-x-2y)/(2x+y). Secondly, set this equal to zero to obtain an expression of y in relation to x or vice versa: x=-2y. Thirdly, input this into the original function to find the x or y coordinate candidates for P: -3y^2+27=0. Therefore the candidates are y=+/-3. Substituting this into either the original equation or the tangent equation will produce candidates for the x-co-ordinate. However, it is easier to insert it into x=-2y. This produces x=-6, or x=6. Given the provided condition, Q(-6,3).

Answered by Rahul P. Maths tutor

7795 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Using the trigonometric identity for tan(A + B), prove that tan(3x)=(3tan(x)-tan^3(x))/(1-3tan^2(x))


Differentiate sin(x^3) with respect to y


How to integrate 5x^2?


The curve has the equation y= (x^3)/(2x-1). Find dy/dx.


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