Tangents and its Equations Test 35

We have,

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

Differentiation this equation with respect to $x$  and we get,

$\dfrac{d}{dx}\left( \sqrt{x}+\sqrt{y} \right)=\dfrac{d}{dx}\sqrt{a}$

$\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$

$\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{x}}$

$\dfrac{dy}{dx}=-\dfrac{\sqrt{y}}{\sqrt{x}}$

At point $\left( {{x}_{1}},{{y}_{1}} \right)$

${{\left( \dfrac{dy}{dx} \right)}_{\left( {{x}_{1}},{{y}_{1}} \right)}}=-\dfrac{\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}}$

 We know that,

Equation of tangent.

$y-{{y}_{1}}=\dfrac{dy}{dx}\left( x-{{x}_{1}} \right)$

At point $\left( {{x}_{1}},{{y}_{1}} \right)$

$y-{{y}_{1}}=-\dfrac{\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}}\left( x-{{x}_{1}} \right)$

$y\sqrt{{{x}_{1}}}-{{y}_{1}}\sqrt{{{x}_{1}}}=-\sqrt{{{y}_{1}}}\left( x-{{x}_{1}} \right)$

$y\sqrt{{{x}_{1}}}-{{y}_{1}}\sqrt{{{x}_{1}}}=-x\sqrt{{{y}_{1}}}+{{x}_{1}}\sqrt{{{y}_{1}}}$

$y\sqrt{{{x}_{1}}}+x\sqrt{{{y}_{1}}}={{y}_{1}}\sqrt{{{x}_{1}}}+{{x}_{1}}\sqrt{{{y}_{1}}}$

On divide $\sqrt{{{x}_{1}}{{y}_{1}}}$ both side and we get,

$\dfrac{y\sqrt{{{x}_{1}}}+x\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}\sqrt{{{y}_{1}}}}=\dfrac{{{y}_{1}}\sqrt{{{x}_{1}}}+{{x}_{1}}\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}\sqrt{{{y}_{1}}}}$

$\dfrac{y}{\sqrt{{{y}_{1}}}}+\dfrac{x}{\sqrt{{{x}_{1}}}}=\sqrt{{{y}_{1}}}+\sqrt{{{x}_{1}}}$

$\dfrac{x}{\sqrt{{{x}_{1}}}}+\dfrac{y}{\sqrt{{{y}_{1}}}}=\sqrt{a}$

We have,

${{x}^{\frac{2}{3}}}+{{y}^{\frac{2}{3}}}={{a}^{\frac{2}{3}}}$

Then,

$\dfrac{d}{dx}\left( {{x}^{\frac{2}{3}}}+{{y}^{\frac{2}{3}}} \right)=\dfrac{d}{dx}\left( {{a}^{\frac{2}{3}}} \right)$

$\dfrac{2}{3}{{x}^{-\frac{1}{3}}}+\dfrac{2}{3}{{y}^{-\frac{1}{3}}}\dfrac{dy}{dx}=0$

${{x}^{-\frac{1}{3}}}+{{y}^{-\frac{1}{3}}}\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}={{\left( -\dfrac{x}{y} \right)}^{-\frac{1}{3}}}$

$\dfrac{dy}{dx}={{\left( -\dfrac{y}{x} \right)}^{\frac{1}{3}}}$

At point $\left( a, 0 \right)$

${{\left( \dfrac{dy}{dx} \right)}_{\left( a,0 \right)}}={{\left( -\dfrac{y}{x} \right)}^{\frac{1}{3}}}_{\left( a,0 \right)}$

${{\left( \dfrac{dy}{dx} \right)}_{\left( a,0 \right)}}={{\left( -\dfrac{0}{a} \right)}^{\frac{1}{3}}}$

${{\left( \dfrac{dy}{dx} \right)}_{\left( a,0 \right)}}=0$

Equation of normal is

$y-{{y}_{1}}=-\dfrac{1}{{{\left( \dfrac{dy}{dx} \right)}_{\left( a,0 \right)}}}\left( x-{{x}_{1}} \right)$

$y-{{y}_{1}}=\dfrac{1}{0}\left( x-{{x}_{1}} \right)$

Equation of normal at point $\left( a,0 \right)$ and we get,

$y-0=\dfrac{1}{0}\left( x-a \right)$

$x-a=0$

$x=a$

Hence, this is the answer.