# This will get you!

1. Aug 31, 2014

### SteliosVas

Here is a good question, been trying to work it out all evening, and were all engineering students, and struggling!!!

Points (p,q) lie on the curve √x + √y = 1

Rearranging to make y the subject we get y=(√x +1)2

We than take the derivative which gives us 1-$\frac{1}{√x}$

Than since we know x=p and y=q

Plugging in p into y' we get 1-$\frac{1}{√p}$

Than into tangent equation y-y1=m(x-x1)

Giving us y - q = 1-$\frac{1}{√p}$ (x-p)

Now compared to the answer which says its √pq or x√q + y√p

I am so lost?

I have attached a picture as well to help!

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2. Aug 31, 2014

### mal4mac

For starters, check that you have rearranged & taken the derivative correctly. Maybe doing dx/dy, as well, will get you somewhere. Just guessing!

3. Aug 31, 2014

### PeroK

You could try Implicit Differentiation.

4. Aug 31, 2014

### PeroK

Something else that might help is:

$$p\sqrt{q} + q\sqrt{p} = \sqrt{pq}(\sqrt{p}+\sqrt{q})$$

5. Aug 31, 2014

### SteliosVas

So I did implict differention of $\sqrt{x}$ + $\sqrt{y}$= 1

and get $\frac{dy}{dx}$=$\frac{-\sqrt{y}}{\sqrt{x}}$

Knowing x=p and y=q

I sub in p (as x) to get m (the gradient of the tangent)

When substituting this into the tangent equation, I still don't get a similar answer.

Am I going wrong somewhere?

Last edited: Aug 31, 2014
6. Aug 31, 2014

### vela

Staff Emeritus
Remember that $\sqrt{p} + \sqrt{q} = 1$.

7. Aug 31, 2014

### SteliosVas

Okay so I have no idea anymore.

I have got the implicit derivative as $\frac{-\sqrt{y}}{x}$.

Therefore the gradient of the tangent to the curve ($\sqrt{x}$+$\sqrt{y}$=1 at points p,q is:

Now as I said earlier, x1=p and y1=q

m= $\frac{-\sqrt{p}}{\sqrt{q}}$

Substituting all known information into the tangent equation of (y-y1)=m(x-x1)

We get y - q = $\frac{-\sqrt{p}}{\sqrt{q}}$ * (x - p)

From here, is where I got no idea?

8. Aug 31, 2014

### SteliosVas

I think this is correct?

Last edited: Sep 1, 2014