Tangents and its Equations Test 51

$$\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right )$$
$$y'(x)= -\cfrac{y^{2/3}}{x^{2/3}}$$
$$\left.y'\right|_{(x,y)=\left(\dfrac{a}{8}, \dfrac{a}{8}\right)} = -1$$

$$4x-3y+2=0$$
Or 
$$3y=4x+2$$
Or 
$$y=\dfrac{4}{3}x+\dfrac{2}{3}$$.
Hence slope of the tangent will be 
$$=\dfrac{-3}{4}$$
Now 
$$\dfrac{dy}{dx}=\dfrac{-3}{4}$$
$$y=\pm\sqrt{2}.x^{\frac{3}{2}}$$
Hence
$$\dfrac{dy}{dx}$$

$$=\pm\sqrt{2}.\dfrac{3}{2}.x^{\frac{1}{2}}$$

$$=\dfrac{-3}{4}$$

Or 
$$-\sqrt{2}.\dfrac{3}{2}.x^{\dfrac{1}{2}}=\dfrac{-3}{4}$$
Or 
$$2\sqrt{2}.x^{\dfrac{1}{2}}=1$$
Or 
$$x=\dfrac{1}{8}$$.
Hence
$$y=-\sqrt{2}.x^{\dfrac{3}{2}}$$
$$=-\sqrt{2}.\dfrac{1}{16\sqrt{2}}$$
$$=-\dfrac{1}{16}$$
Hence the co-ordinates are  $$(\dfrac{1}{8},\dfrac{-1}{16})$$

Given, $$\displaystyle f\left ( x \right )=x^{4}-6x^{3}+12x^{2}-8x+3$$
At inflection point $$ f''(x) = 0 $$
$$ 12x^2-36x+24 = 0$$
$$(x-2)(x-1) = 0$$
$$x =2,1$$
$$f'(x)  = 4x^3-18x^2+24x-8$$
$$f'(1) = 2$$ and $$f'(2) = 0$$
$$f(1) = 2$$ and $$f(2) = 3 $$
So the equations are $$ y=3 $$ and $$ y=2x$$

$$3y^{2}.y'+6x=12y'$$
Or 
$$y'(3y^{2}-12)=-6x$$
Or 
$$y'(y^{2}-4)=-2x$$
Or 
$$y'=\dfrac{-2x}{y^{2}-4}$$
Hence for the tangent to be vertical
$$y^{2}-4=0$$
$$y=\pm2$$
Substituting $$y=2$$, we get 
$$3x^{2}=12y-y^{3}$$
$$=24-8$$
$$=16$$
Hence
$$x=\dfrac{\pm4}{\sqrt{3}}$$
$$y=-2$$ gives 
$$3x^{2}=-16$$ 
This is not possible.
Hence the required point is 
$$(\dfrac{\pm4}{\sqrt{3}},2)$$.

Given, $$2y+5x=10$$
$$2y'+5=0$$
$$y'=\dfrac{-5}{2}$$.
Hence the required slope is $$\dfrac{2}{5}$$.
Now 
$$4x^{2}-9y^{2}=36$$.
$$4x^{2}-36=9y^{2}$$
$$y=\dfrac{1}{3}.\sqrt{4x^{2}-36}$$.
$$y'=\dfrac{8x}{6\sqrt{4x^{2}-36}}$$ $$=\dfrac{2}{5}$$
$$40x=12\sqrt{4x^{2}-36}$$
$$10x=3\sqrt{4x^{2}-36}$$
$$100x^{2}=36x^{2}-324$$
$$64x^{2}=-324$$
$$x^{2}=\dfrac{-81}{16}$$.
Hence no real values of x.
Hence answer is none of these.

Slope of the tangents will be 
$$y'=\dfrac{-12x}{(x^{2}+3)^{2}}$$
Now for maximum and minimum slopes
$$y"=0$$
Or 
$$-12(x^{2}+3)^{2}+24x(x^{2}+3).2x=0$$
Or 
$$(x^{2}+3)[48x^{2}-12x^{2}-36]=0$$
Or 
$$36x^{2}-36=0$$
Or 
$$x=\pm1$$.
Hence
$$y'(1)=\dfrac{-12}{16}=\dfrac{-3}{4}$$
And 
$$y'(-1)=\dfrac{3}{4}$$.
Hence $$y(1)=y(-1)=\dfrac{6}{4}=\dfrac{3}{2}$$.
Hence the equations are 
$$y-\dfrac{3}{2}=\dfrac{3}{4}(x+1)$$
Or 
$$4y-6=3x+3$$
Or 
$$3x-4y+9=0$$
And 
$$y-\dfrac{3}{2}=\dfrac{-3}{4}(x-1)$$
Or 
$$4y-6=-3x+3$$
Or 
$$3x+4y-6=0$$.

It is given that 
$$\dfrac{df(x)}{dx}=f(x)^{2}$$
Let 
$$f(x)=y$$
Then
$$y'=y^{2}$$
Or 
$$\int y^{-2}.dy=\int dx$$
Or 
$$\dfrac{-1}{y}=x+c$$
Now 
$$y=\dfrac{-1}{2}$$ at x=0.
Hence
$$c=2$$
Or 
$$-\dfrac{1}{y}=x+2$$
Or 
$$xy+2y+1=0$$ ...(i)
$$\dfrac{dy}{dx}_{x=0}$$
$$=y^{2}_{y=\frac{-1}{2}}$$
Or 
$$y'=\dfrac{1}{4}$$.
Now equation of the tangent is 
$$y-(\dfrac{-1}{2})=\dfrac{1}{4}(x-0)$$
Or 
$$4y+2=x$$
Or 
$$x-4y=2$$ is the required equation of tangent.

Any point on curve $$\displaystyle y=x^{2}$$ is $$\displaystyle P\left ( t,t^{2} \right )$$
$$\displaystyle

\frac{dy}{dx}=2x$$ equation of normal at $$\displaystyle \left (

t,t^{2} \right )$$ is$$\displaystyle y-t^{2}=-\frac{1}{2t}\left ( x-t

\right )$$
solving with $$\displaystyle y=x^{2}$$ we get
$$\displaystyle

x^{2}-t^{2}=\frac{-1}{2t}\left ( x-t \right )\Rightarrow \left ( x-t

\right )\left ( x+t+\frac{1}{2t} \right )=0\Rightarrow

x=-t-\frac{1}{2t}$$
So normal cuts the curve again at $$\displaystyle Q\left ( -t-\frac{1}{2t},\left ( -t-\frac{1}{2t} \right )^{2} \right )$$
$$\displaystyle z=PQ^{2}=4t^{2}\left ( 1+\frac{1}{4t^{2}} \right )^{3}$$
$$\displaystyle

\frac{dz}{dt}=0\Rightarrow t=\pm \frac{1}{\sqrt{2}}=0,\frac{dz}{dt}$$

changes sign from negative to positive about $$\displaystyle

t=\frac{1}{\sqrt{2}}$$ as well as $$\displaystyle

t=-\frac{1}{\sqrt{2}}$$(No chord is format for t=0)
z is minimum

at $$\displaystyle t=\pm \frac{1}{\sqrt{2}}$$ & minimum value

of $$\displaystyle z=PQ^{2}=3$$ shortest normal chord has

length $$\displaystyle \sqrt{3}$$ & its equation is $$\displaystyle

x+\sqrt{2y}-\sqrt{2}=0$$ or $$\displaystyle x-\sqrt{2y}+\sqrt{2}=0$$