Explanation
Equation of tangent is y-2x+1=0
It is tangent at x=1, so for x=1
y-2(1)+1=0\\ \Rightarrow y=1
So, its is tangent to the curve at (1,1)
Slope of tangent = -\left (\dfrac{-2}{1}\right)=2
xy+ax+by=0\\ y+x\dfrac { dy }{ dx } +a+b\dfrac { dy }{ dx } =0
Now \dfrac { dy }{ dx } =2 at (1,1)
1+1(2)+a+b(2)=0\\ \Rightarrow a+2b=-3 .....(i)
(1,1) also lies on the curve xy+ax+by=0
\Rightarrow 1(1)+a(1)+b(1)=0\\ \Rightarrow a+b=-1 .......(ii)
Solving (i) and (ii), we get
\Rightarrow a=1,b=-2
So, option E is correct.
\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}\,\,\,\,\,\,\,\left( {given} \right)
\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{{dx}}{{{x^2}}}} }
\Rightarrow \log y = - \dfrac{1}{x} + c
\Rightarrow y = {e^{ - \frac{1}{x} + c}}
\Rightarrow y = {e^{ - \frac{1}{x}}},c
since this curve is passing through point \left( {1,3} \right)
so, 3 = {e^{ - \dfrac{1}{t}}}.c
\Rightarrow c = 3e
therefore , eqn of the curve is
y = {e^{ - \dfrac{1}{x}}}.3e
\Rightarrow y = 3{e^{ - \dfrac{1}{x} + 1}}
\Rightarrow y = 3{e^{1 - \dfrac{1}{x}}}
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)
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}
{{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.
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