Finding equation of tangent line

  • Thread starter Dank2
  • Start date
  • #26
213
4
Sort of, although what you wrote isn't clear. You can't just replace y with x or x with y, but you can replace y with ##(1 - \sqrt{x})^2)## and you can replace ##y_0## with ##(1 - \sqrt{x_0})^2)##.

Your work would be easier to read if you used LaTeX. If you click on any of the expressions I wrote using LaTeX, you can see my script that produces it.

For example, this -- ##\sqrt{y_0}## renders as ##\sqrt{y_0}##.

BTW, I'm wondering if there's a mistake in the problem statement. I am getting ##\frac{y}{\sqrt{y_0}} + \frac{x}{\sqrt{x_0}} = 2##, not 1 as in post #1. It's possible I have an error.

The problem would be slightly easier if instead of the point ##(x_0, y_0)## we use the point (a, b).

I will try to use latex once I get to my pc.

Yes I know that, I should have said y in terms of x or y_0 in terms of x_0.

So the right way is trying all possibilities ? Or is there some method I should follow?

And it should equal to 1.
 
  • #27
213
4
Sort of, although what you wrote isn't clear. You can't just replace y with x or x with y, but you can replace y with ##(1 - \sqrt{x})^2)## and you can replace ##y_0## with ##(1 - \sqrt{x_0})^2)##.

Your work would be easier to read if you used LaTeX. If you click on any of the expressions I wrote using LaTeX, you can see my script that produces it.

For example, this -- ##\sqrt{y_0}## renders as ##\sqrt{y_0}##.

BTW, I'm wondering if there's a mistake in the problem statement. I am getting ##\frac{y}{\sqrt{y_0}} + \frac{x}{\sqrt{x_0}} = 2##, not 1 as in post #1. It's possible I have an error.

The problem would be slightly easier if instead of the point ##(x_0, y_0)## we use the point (a, b).

Ok I've just solved it taking the slope/point equation and the the other , putting both in terms of y, making the nominator equal simplifying . Thanks
 

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