Cartesian eq- help with this q pls ??

Oors

New member
Joined
Mar 28, 2020
Messages
7
Screenshot 2020-03-28 at 18.03.19.png
really struggling with part b and c
part A I calculated t= +/- 3
 
Hello, and welcome to FMH! :)

For part (b) consider:

[MATH]x^2=t^2+6+\frac{9}{t^2}[/MATH]
[MATH]y^2=t^2-6+\frac{9}{t^2}[/MATH]
What do you get upon subtracting the latter from the former?
 
Hello, and welcome to FMH! :)

For part (b) consider:

[MATH]x^2=t^2+6+\frac{9}{t^2}[/MATH]
[MATH]y^2=t^2-6+\frac{9}{t^2}[/MATH]
What do you get upon subtracting the latter from the former?
Thank you I got the answer !
 
Hello, and welcome to FMH! :)

For part (b) consider:

[MATH]x^2=t^2+6+\frac{9}{t^2}[/MATH]
[MATH]y^2=t^2-6+\frac{9}{t^2}[/MATH]
What do you get upon subtracting the latter from the former?
How do you obtain an answer for part c ?? :)
 
Find the slope at t=1. Then what would the slope of the normal line be?
 
Find the slope at t=1. Then what would the slope of the normal line be?
got answer y+2=1/2(x-4) for the equation of the line however I am stuck on how to do part (ii)
 
Okay, for part (b) you should have found that:

[MATH]x^2-y^2=12[/MATH]
The slope of the normal lime is:

[MATH]m=\left.-\frac{dx}{dy}\right|_{t=1}=\frac{1}{2}[/MATH]
And so the normal line to \(P\) is:

[MATH]y=\frac{1}{2}(x-4)-2=\frac{1}{2}x-4[/MATH]
This is equivalent to your result. To find the other point of intersection, we can substitute for \(y\):

[MATH]x^2-\left(\frac{1}{2}x-4\right)^2=12[/MATH]
What do you get when you solve the above for \(x\)?

fmh_0132.png
 
Top