Tangents and its Equations Test 19

Hence slope of tangents
$$=tan60^{0}$$
$$=\sqrt{3}$$
$$=\dfrac{dy}{dx}$$
Hence
$$\dfrac{dy}{dx}$$
$$=3x^{2}$$
$$=\sqrt{3}$$
Or 
$$x^{2}=\dfrac{1}{\sqrt{3}}$$
Hence
$$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}$$.
Thus y
$$=\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$$.

$$\dfrac{dy}{dx}=tan45^{0}$$
Or
$$\dfrac{dy}{dx}=1$$
Or 
$$\dfrac{(1-x^{2})-x(-2x)}{(1-x^{2})^{2}}=1$$
Or 
$$1-x^{2}+2x^{2}=(1-x^{2})^{2}$$
Or 
$$1+x^{2}=1-2x^{2}+x^{4}$$
Or 
$$x^{4}-3x^{2}=0$$
Or 
$$x^{2}[x^{2}-3]=0$$
$$x=0$$ or 
$$x=\pm\sqrt{3}$$
Hence
$$y=\dfrac{\sqrt{3}}{1-3}=-\dfrac{\sqrt{3}}{2}$$

$$y=\dfrac{-\sqrt{3}}{1-3}=\dfrac{\sqrt{3}}{2}$$

Hence
$$\left(\sqrt{3},-\dfrac{\sqrt{3}}{2}\right),\left(-\sqrt{3},\dfrac{\sqrt{3}}{2}\right)$$.

Let the equation of curve be $$y=x^4$$
Let the point of contact be $$(h,k)$$
$$\Rightarrow \displaystyle k= h^{4}$$    ...(1)

Now, $$\dfrac{dy}{dx}=4x^3$$
Slope of tangent to the curve at $$(h,k)$$ is $$4h^3$$
Tangent is $$\displaystyle y-k= 4h^{3}\left ( x-h \right )$$ ...(2)
It passes through (2,0) 
$$\displaystyle \therefore -k= 4h^{3}\left ( 2-h \right )$$ 
$$\displaystyle -h^{4}= 8h^{3}-4h^{4}$$ by (1)
$$\displaystyle 3h^{4}-8h^{3}= 0$$
$$\displaystyle \therefore h= 0$$ or $$\dfrac{8}{3} $$
$$\displaystyle \therefore k= 0$$ or $$\displaystyle \left ( 8/3 \right )^{4}$$
$$\displaystyle \therefore $$ Points are (0,0) and $$\displaystyle \left [ 8/3, \left ( 8/3 \right )^{4} \right ]$$ 
Putting in (2), tangents are $$\displaystyle y= 0$$
and $$\displaystyle y-\left ( \frac{8}{3} \right )^{4}= 4\left ( \frac{8}{3} \right )^{3}\left ( x-\frac{8}{3} \right )$$

Let the tangent be drawn at the point $$\left(x,y\right).$$

The equation of the curve does not change if $$x$$ be + ive or-ive. 
Hence we consider only +ive values of $$x$$ i.e $$\displaystyle x> 0,$$ 
$$\displaystyle \therefore \left | x \right |=x \therefore y={-x}$$
 Consider any point (x,y) tangent at which is $$\displaystyle Y-y=-e^{-x}\left ( X-x \right )$$ 
It meets the axis $$\displaystyle Y=0 at \left ( x=ye^{x},0 \right )$$ 
It meets the axis $$\displaystyle X=0 at (0,y+\dfrac x{e^{x}})$$ 
Area of the triangle $$\displaystyle =\frac{1}{2}AB $$
 $$\displaystyle \Delta =\frac{1}{2}\left ( x=ye^{x} \right )\left ( y+\frac{x}{e^{x}} \right )$$ 
$$\displaystyle =\frac{1}{2}\left ( xy+xy+y^{2}e^{x}+\frac{x^{2}}{e^{x}} \right )$$ Substitute $$\displaystyle y=\frac{1}{e^{x}}$$ 
$$\displaystyle \Delta =\frac{1}{2}\left [ \frac{2x}{e^{x}}+\frac{e^{x}}{e^{2x}}+\frac{x^{2}}{e^{x}} \right ]=\frac{1}{2}e^{-x}\left ( x+1 \right )^{2}$$ 
For max. value of $$\displaystyle \Delta $$
 we have$$\displaystyle \frac{d\Delta }{dx}=0 \therefore \frac{1}{2}e^{-x\left [ -\left ( x+1 \right )^{2}+2\left ( x+1 \right ) \right ]}=0 $$ 
or $$\displaystyle \frac{1}{2}\left ( x+1 \right )\left ( 1-x \right )e^{-x}=\frac{1}{2}e^{-x}\left ( 1-x^{2} \right )$$ 
$$\displaystyle \therefore \frac{d\Delta }{dx}=0\Rightarrow x=1-1.$$
We shall consider only x=1as x is +ive.
$$\displaystyle \frac{d^{2}\Delta }{dx^{2}}=\frac{1}{2}e^{x}\left [ -1\left ( 1-x^{2} \right )-2x \right ]$$ 
$$\displaystyle =-\frac{x}{e^{x}}=-\frac{1}{e}=-ive$$ at $$x=1 $$
 Hence $$\displaystyle \Delta $$ is max.
When x=1 $$\displaystyle \therefore y=e^{-1}=\dfrac 1e.$$ 
$$\displaystyle \therefore $$ 
Required point is $$(1,\dfrac 1e)$$. we have already stated that if $$x$$ be changed to $$-x$$ the equation of the curve does not change. 
Hence the required points are $$\displaystyle \left ( \pm 1,\dfrac 1e \right )$$ 

Given, $$ay^2 = x^3$$

Given equation of parabola is 
$$\displaystyle y= ax^{2}+bx+c$$ 
$$\displaystyle \frac{dy}{dx}= 2ax+b$$
Slope of tangent to the curve at $$x=1$$ is $$2a+b$$
Given tangent is $$y=x$$ . Slope of this tangent is $$1.$$
So, $$2a+b=1$$    ...(1)

Since, the parabola passes through $$(-1,0)$$ 
$$\displaystyle \therefore a-b+c= = 0$$      ...(2) 

Given $$y=x$$ is a tangent at $$x=1$$
 $$\displaystyle \therefore y= 1.$$ 
Hence $$(1,1)$$ lies both on tangent and parabola 
$$\displaystyle \therefore a+b+c= 1$$         ...(3)

Solving (1), (2) and (3), we get
 $$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$$