Cartesian eq- help with this q pls ??

Oors

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Screenshot 2020-03-28 at 18.03.19.png
really struggling with part b and c
part A I calculated t= +/- 3
 

pka

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View attachment 17489
really struggling with part b and c
part A I calculated t= +/- 3
Can you find both \(\dfrac{dx}{dt}~\&~\dfrac{dy}{dt}~?\) If yes then \(\dfrac{dx}{dy}=\dfrac{dx/dt}{dy/dt}\).
If no what have you been doing in the course?
 

MarkFL

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Hello, and welcome to FMH! :)

For part (b) consider:

\(\displaystyle x^2=t^2+6+\frac{9}{t^2}\)

\(\displaystyle y^2=t^2-6+\frac{9}{t^2}\)

What do you get upon subtracting the latter from the former?
 

Oors

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Hello, and welcome to FMH! :)

For part (b) consider:

\(\displaystyle x^2=t^2+6+\frac{9}{t^2}\)

\(\displaystyle y^2=t^2-6+\frac{9}{t^2}\)

What do you get upon subtracting the latter from the former?
Thank you I got the answer !
 

Oors

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Hello, and welcome to FMH! :)

For part (b) consider:

\(\displaystyle x^2=t^2+6+\frac{9}{t^2}\)

\(\displaystyle y^2=t^2-6+\frac{9}{t^2}\)

What do you get upon subtracting the latter from the former?
How do you obtain an answer for part c ?? :)
 

Jomo

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Find the slope at t=1. Then what would the slope of the normal line be?
 

Oors

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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)
 

MarkFL

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Okay, for part (b) you should have found that:

\(\displaystyle x^2-y^2=12\)

The slope of the normal lime is:

\(\displaystyle m=\left.-\frac{dx}{dy}\right|_{t=1}=\frac{1}{2}\)

And so the normal line to \(P\) is:

\(\displaystyle y=\frac{1}{2}(x-4)-2=\frac{1}{2}x-4\)

This is equivalent to your result. To find the other point of intersection, we can substitute for \(y\):

\(\displaystyle x^2-\left(\frac{1}{2}x-4\right)^2=12\)

What do you get when you solve the above for \(x\)?

fmh_0132.png
 
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