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)

LB
Answered by Luca B. Further Mathematics tutor

2920 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Let E be an ellipse with equation (x/3)^2 + (y/4)^2 = 1. Find the equation of the tangent to E at the point P where x = √3 and y > 0, in the form ax + by = c, where a, b and c are rational.


How do I find and plot the roots of a polynomial with complex roots on an Argand diagram? e.g. f(z) =z^3 -3z^2 + z + 5 where one of the roots is known to be 2+i


Why is the integral of 1/sqrt(1-x^2)dx = sin^{-1}(x)?


I do not understand this topic and particularly this example. In the class the result was found out but I still do not get it. How did the teacher came up with this outcome?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences