Find the equation of the tangent to the curve

Question

Find the equation of the tangent to the curve $\sqrt X + \sqrt Y = a\;$ at the point $\left(\dfrac {a^2}{4},\dfrac {a^2}{4}\right)$

I don't know how to find $\dfrac {\mathrm dy}{\mathrm dx}$ in this particular question... Please help me to do this sum.

Answer 1

HINT:

$$\sqrt x+\sqrt y=a$$

Differentiating with respect to $x, \frac 1{2\sqrt x}+\frac1{2\sqrt y}\cdot \frac{dy}{dx}=0$ as $\frac{d g(y)}{dx}=\frac{d g(y)}{dy}\cdot\frac{dy}{dx}$

At $x=y=\frac{a^2}4,$ $$\frac1a+\frac1a\cdot \frac{dy}{dx}=0\implies \frac{dy}{dx}=-1$$ (assuming $a\ne0$)

Answer 2

$$\sqrt x+\sqrt y=a$$ $$ y=(a-\sqrt x)^2$$ $$ y=a^2+x-2a\sqrt x$$ $$\dfrac {dy}{dx}=0+1-2a\dfrac {1}{2\sqrt x}$$ $$\dfrac {dy}{dx}=1-\dfrac {a}{\sqrt x}$$ put $x=\dfrac {a^2}{4}$ $$\dfrac {dy}{dx}_{x=\frac {a^2}{4}}=1-\dfrac {a}{\sqrt{\frac {a^2}{4}}}$$ $$\dfrac {dy}{dx}=-1$$

Answer 3

You do not even need to differentiate, or know anything about derivatives. The curve is symmetrical in $x$ and $y$, and the point we are interested in also does not change when we interchange $x$ and $y$. So the slope must be $1$ or $-1$. A quick sketch rules out the possibility that it is $1$.