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 ▸

Given that x = i is a solution of 2x^3 + 3x^2 = -2x + -3, find all the possible solutions


How do I integrate (sin x)^6?


Give the general solution to the Ordinary Differential Equation: (dy/dx) + 2y/x = 3x+2


Find the general solution to the differential equation: d^2y/dx^2 - 8 dy/dx +16y = 2x


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