Cartesian eq- help with this q pls ??
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pka
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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:
[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 ??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?
got answer y+2=1/2(x-4) for the equation of the line however I am stuck on how to do part (ii)
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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 \(x\)?
[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 \(x\)?