The rectangular hyperbola H has parametric equations: x = 4t, y = 4/t where t is not = 0. The points P and Q on this hyperbola have parameters t = 1/4 and t = 2 respectively. The line l passes through the origin O and is perpendicular to the line PQ.

This question asks us to find the cartesian equation of l.
First we must find the points P and Q. To do this we substitute t with 1/4 to find P and substitute t with 2 to find Q.Doing this we get the coordinates for P and Q.For P: x= 4(1/4) = 1 y= 4/(1/4) = 16 P(1,16)For Q: x = 4(2) = 8 y = 4/2 = 2 Q(8,2)
The equation of line l is found by using the standard method y - y* = m(x - x*) where y* and x* are points on the line and m is the gradient.We must find the gradient of PQ before we find the gradient of l. To do this we simply use dy/dx: (Gradient PQ) = (16-2)/(1-8) = -(14/7) = -2
As PQ is perpendicular to l, we follow this formula. (Gradient of l) * (Gradient PQ) = -1: Gradient of l must be 1/2 as gradient of l = -1/(Gradient PQ)
We can now use y - y* = m(x - x*). We know that l passes through the origin, and therefore x* = 0 and y* = 0: y - 0 = 1/2(x - 0) y = 1/2x (this is the equation of l, as required)

Related Further Mathematics A Level answers

All answers ▸

How do I find the vector/cross product of two three-dimensional vectors?


using an integrating factor, find the general solution of the differential equation dy/dx +y(tanx)=tan^3(x)sec(x)


Let A, B and C be nxn matrices such that A=BC-CB. Show that the trace of A (denoted Tr(A)) is 0, where the trace of an nxn matrix is defined as the sum of the entries along the leading diagonal.


What modules have you done before?


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 2013–2024

Terms & Conditions|Privacy Policy