How to solve this question?

lollipop

New member
Joined
Oct 7, 2014
Messages
6
Please help me to solve question 2
 

Attachments

  • image.jpg
    image.jpg
    260.2 KB · Views: 10
Not only is it extremely hard to read (for one thing, the image is on its side) but part of the first line appears to be cut off.
I think the question is "if a single line passes through P, Q, R, and S, four points on the hyperbola x= a sec(t), y= b tan(t), then |PR|= |QS|". Is that correct?
 
Last edited:
I think the question is "if a single line passes through P, Q, R, and S, four points on the hyperbola x= a sec(t), y= b tan(t), then |PR|= |QS|". Is that correct?
You're close:

If the lines passing through \(\displaystyle P\) and \(\displaystyle Q\) on the hyperbola \(\displaystyle \begin{cases}x\, =\, a\sec\left(\theta\right)\\y\, =\, b\tan\left(\theta\right)\end{cases}\) intersect with the asymptotes at \(\displaystyle R\) and \(\displaystyle S\) respectively, prove that \(\displaystyle \left|PR\right|\, =\, \left|QS\right|\)

You're quite right: That image is a bear to try to read! :shock:
 
The first thing I would do is change to Cartesian coordinates: since \(\displaystyle x= a sec(theta)\) and \(\displaystyle y= b tan(\theta)\)
\(\displaystyle \frac{x^2}{a^2}- \frac{y^2}{b^2}= sec^2(\theta)- tan^2(\theta)= 1\). That is a hyperbola with center at (0, 0). For x and y very large compared with a and b, "1" is very small compared with the squares so, approximately, \(\displaystyle \frac{x^2}{a^2}- \frac{y^2}{b^2}= 0\) which is the same as \(\displaystyle \frac{x}{a}= \pm\frac{y}{b}\). Its asymptotes are \(\displaystyle y= bx/a\) and \(\displaystyle y= -bx/a\). Let P and Q be points on the hyperbola and write the equations of possible lines through them. Where do those cross the asymptotes?
 
How to prove X3+X4=asec theta1 + asec theta2 ?
 

Attachments

  • image.jpg
    image.jpg
    336.2 KB · Views: 5
Okay; there's no way in heck I'm gonna try to read that sideways graphic! :shock:
 
Top