Question:
Find the equation of the tangent to the curve $\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=\mathrm{a}$, at the point $\left(\mathrm{a}^{2} / 4, \mathrm{a}^{2} / 4\right)$
Solution:
finding slope of the tangent by differentiating the curve
$\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}}\left(\frac{d y}{d x}\right)=0$
$\frac{d y}{d x}=-\frac{\sqrt{x}}{\sqrt{y}}$
at $\left(\frac{\mathrm{a}^{2}}{4}, \frac{\mathrm{a}^{2}}{4}\right)$ slope $\mathrm{m}$, is $-1$
the equation of the tangent is given by $y-y_{1}=m\left(x-x_{1}\right)$
$\mathrm{y}-\frac{\mathrm{a}^{2}}{4}=-1\left(\mathrm{x}-\frac{\mathrm{a}^{2}}{4}\right)$
$x+y=\frac{a^{2}}{2}$
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