CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 8 - MCQExams.com

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.

$$\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}\,\,\,\,\,\,\,\left( {given} \right)$$

$$\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{{dx}}{{{x^2}}}} } $$

$$ \Rightarrow \log y =  - \dfrac{1}{x} + c$$

$$ \Rightarrow y = {e^{ - \frac{1}{x} + c}}$$

$$ \Rightarrow y = {e^{ - \frac{1}{x} + c}}$$

$$ \Rightarrow y = {e^{ - \frac{1}{x}}},c$$

 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.

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

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.