Let (p,q) be a point on the curve sqrt(x) + sqrt(y)=1 and dy/dx=-sqrt(y/x). Show that the tangent line to this curve at (p,q) has equation (x)sqrt(q) + (y)sqrt(p) = sqrt (pq).

I can't seem to get the correct answer for the RHS.

Question

Let (p,q) be a point on the curve sqrt(x) + sqrt(y)=1 and dy/dx=-sqrt(y/x). Show that the tangent line to this curve at (p,q) has equation (x)sqrt(q) + (y)sqrt(p) = sqrt (pq).

I can't seem to get the correct answer for the RHS.

anonymous · Answer

This is what I keep getting: $$x \sqrt{q} + y \sqrt{p} = p \sqrt{q} + q \sqrt{p}$$

ganeshie8 · Answer

Point :  $\large(p, q)$
Slope :  $\large -\frac{q}{p}$ 

Line :
$\large  y - q  =   -\sqrt{\frac{q}{p}}(x-p)$

anonymous · Answer

yep, I did that

ganeshie8 · Answer

simplifying gives  :

$\large  x\sqrt{q} + y \sqrt{p} =  q\sqrt{p} +   \sqrt{pq}$

anonymous · Answer

the answer requires only $$\sqrt{pq}$$ :/

ganeshie8 · Answer

yeah, not sure how to kick off that extra $\large q\sqrt{p}$ from right hand side :/

anonymous · Answer

damn.. thanks anyway :)

ganeshie8 · Answer

np :) its not wise to call the text book has a mistake... but.. im not getting any other clue  ...

anonymous · Answer

it's actually an assignment :P

anonymous · Answer

Oh, I've got it. Take $$\sqrt{pq}$$ out as a common factor so you're left with: $$\sqrt{pq}(\sqrt{p} + \sqrt{q})$$ Sub (p,q) into the equation given to so that: $$\sqrt{p} + \sqrt{q} = 1$$ And the RHS will equal: $$\sqrt{pq}$$

ganeshie8 · Answer

brilliant !! xD