1. ## Tangent & Normal(Q6)

Given that $\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-x)^2}$ for the curve $y=\frac{x}{\sqrt{x}-2}$, obtain the equation of the normal to the curve $y=\frac{x}{\sqrt{x}-2}$ at the point on the curve where x=1.

I have attempted the question by finding the value of y when x=1. Then using the formula $(y-y_1)=\frac{dy}{dx}(x-x_1)$. However, i am unable to simplify it...

Hence, i require a detailed working to examine the problem, thanks!

2. Originally Posted by Punch
Given that $\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-\color{red}2)^2}$ for the curve $y=\frac{x}{\sqrt{x}-2}$, obtain the equation of the normal to the curve $y=\frac{x}{\sqrt{x}-2}$ at the point on the curve where x=1.

I have attempted the question by finding the value of y when x=1. Then using the formula $(y-y_1)=\frac{dy}{dx}(x-x_1)$. However, i am unable to simplify it...

Hence, i require a detailed working to examine the problem, thanks!
i think you have a typo (red) , when x=1 , dy/dx=-3/2 , so gradient of normal=2/3

the normal cuts the curve at (1,-1)

equation : y+1=(2/3)(x-1)

3. Hi, mathaddict, please note that there isn't any typo. I have already tried differentiating the curve and proved it to be $
\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-2)^2}
$

4. Originally Posted by Punch
Hi, mathaddict, please note that there isn't any typo. I have already tried differentiating the curve and proved it to be $
\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-2)^2}
$

$\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-x)^2}$ , this is what you wrote initially , then i tried differentiating myself and hv checked with some calculus engines, i got this one

$\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-2)^2}$

$\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-x)^2}$ , this is what you wrote initially , then i tried differentiating myself and hv checked with some calculus engines, i got this one
$\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-2)^2}$