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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?

Thank you I got the answer !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 ??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?

got answer y+2=1/2(x-4) for the equation of the line however I am stuck on how to do part (ii)Find the slope at t=1. Then what would the slope of the normal line be?

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

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 \(x\)?