Tangents and its Equations Test 18

$$\dfrac{dy}{dx}+e^{xy}[x\dfrac{dy}{dx}+y]+1=0$$
Or 
$$\dfrac{dy}{dx}[1+x.e^{xy}]+1+y.e^{xy}=0$$
Or 
$$\dfrac{dy}{dx}=\dfrac{-(1+y.e^{xy})}{1+x.e^{xy}}$$
Now 
$$1+xe^{xy}=0$$ since the slope of the tangent is $$90^{0}$$.
Hence
$$xe^{xy}=-1$$
Or 
$$x(-x-y)=-1$$
Or 
$$x^{2}+xy=1$$
Or 
$$y=\dfrac{1-x^{2}}{x}$$
Or 
$$y=\dfrac{1}{x}-x$$ ...(i)
Considering $$y=0$$,
$$x^{2}=1$$
$$x=\pm1$$.
Hence we get 2 points $$(1,0)$$ and $$(-1,0)$$.
Out of these 2, only $$(-1,0)$$ lies on the curve.
Hence the required point is $$(-1,0)$$.

$$dx=a\sin \theta.d\theta$$
$$dy=a\cos \theta.d\theta$$
Or 
$$\dfrac{-dx}{dy}=-\tan\theta$$.
Thus
Equation of normal will be 
$$y-a\sin \theta=-\tan\theta(x-a+a\cos \theta)$$
Or 
$$\cos \theta.y-a\sin \theta.\cos \theta=-x\sin \theta+a\sin \theta-a\sin \theta \cos \theta$$
$$\cos \theta.y+\sin \theta.x=a\sin \theta$$ 
Hence it passes through the fixed point $$(a,0)$$.

Slope of normal
$$=\dfrac{-1}{f'(x)}$$
$$=tan(\dfrac{3\pi}{4})$$
$$=-1$$
Hence
$$f'(x)=1$$
Or 
$$\dfrac{dy}{dx}=1$$
Or 
$$y=x+c$$
Hence
$$y=f(x)$$ is an equation of a straight line parallel to $$y=x$$.
Hence
$$f'(x)$$ is independent of x and its value is 1.

Given, $$\displaystyle \frac{x^{m}}{a^{m}}+\frac{y^{m}}{b^{m}}=1.$$  ..(1)
Differentiating w.r.t. x, we get 
$$\displaystyle m.\frac{x^{m-1}}{a^{m}}+m.\frac{y^{m-1}}{b^{m}}\cdot \frac{dy}{dx}=0.$$
$$\displaystyle \therefore\frac{dy}{dx} =\frac{1}{a}\left ( \frac{x}{a} \right )^{m-1}.b\left ( \frac{b}{y} \right )^{m-1}$$
$$\displaystyle \therefore$$ Equation of tangent is $$\displaystyle Y-y=\frac{dy}{dx}\left ( X-x \right )$$
or $$\displaystyle Y-y=-\frac{1}{a}\left ( \frac{x}{a} \right )^{m-1}.b\left ( \frac{b}{y} \right )^{m-1}\left ( X-x \right )$$
or $$\displaystyle

\frac{X}{a}\left ( \frac{x}{a} \right )^{m-1}-\left ( \frac{x}{a}

\right )^{m}=-\frac{Y}{b}\left ( \frac{y}{b} \right )^{m-1}+\left (

\frac{y}{b} \right )^{m}$$
or $$\displaystyle \frac{X}{a}\left (

\frac{x}{a} \right )^{m-1}+\frac{Y}{b}\left ( \frac{y}{b} \right

)^{m-1}=\left ( \frac{x}{a} \right )^{m}+\left ( \frac{y}{b} \right

)^{m}$$
or $$\displaystyle \frac{X}{a}\left ( \frac{x}{a} \right )^{m-1}+\frac{Y}{b}\left ( \frac{y}{b} \right )^{m-1}=1,$$ by (1)

$$x^{2/3}+y^{2/3}=a^{2/3}$$

Given $$x=\displaystyle \frac{2at^{2}}{1+t^{2}}$$, $$ y= \displaystyle \frac{2at^{3}}{1+t^{2}}$$
At $$t=\dfrac{1}{2}$$ we get
$$x=\dfrac{2a}{5}$$ , $$y=\dfrac{a}{5}$$
Hence, the point is $$(\dfrac{2a}{5},\dfrac{a}{5})$$

Now, $$\dfrac { dx }{ dt } =2a\dfrac { (1+t^{ 2 })2t-{ 2t }^{ 3 } }{ { (1+t^{ 2 }) }^{ 2 } } =\dfrac { 4at }{ { (1+t^{ 2 }) }^{ 2 } } $$

$$\dfrac { dy }{ dt } =2a\dfrac { (1+t^{ 2 })3t^{ 2 }-{ t }^{ 3 }(2t) }{ { (1+t^{ 2 }) }^{ 2 } } =\dfrac { 2a(3t^{ 2 }+{ t }^{ 4 }) }{ { (1+t^{ 2 }) }^{ 2 } } $$

Now, $$\dfrac{dy}{dx}=\dfrac { (3t^{ 2 }+{ t }^{ 4 }) }{ 2t } $$

Slope of tangent at $$t=\dfrac{1}{2}$$ is $$\dfrac { 13 }{ 16a } $$

Equation of tangent at $$(\dfrac{2a}{5},\dfrac{a}{5})$$ is 
$$y-\dfrac{a}{5}=\dfrac { 13 }{ 16a } (x-\dfrac{2a}{5})$$
$$\Rightarrow 13x-16y=2a$$

Equation of normal at $$(\dfrac{2a}{5},\dfrac{a}{5})$$ is 
$$y-\dfrac{a}{5}=-\dfrac { 16a }{ 13 } (x-\dfrac{2a}{5})$$
$$\Rightarrow 16x+13y=9a$$

Substituting $$x=2$$ in the equation, we get 
$$y=8-8+4=4$$
Thus the point is $$(2,4)$$.
Now 
$$dy=3x^{2}-4x.dx$$
Or 
$$\dfrac{-dx}{dy}=\dfrac{-1}{3x^{2}-4x}$$
Now 
$$\dfrac{-dx}{dy}_{x=2}=\dfrac{-1}{12-8}$$

$$=\dfrac{1}{4}$$.
Hence equation of normal will be 
$$y-4=(x-2).\dfrac{-1}{4}$$
Or 
$$4y-16=2-x$$
Or 
$$x+4y=18$$