Tangents and its Equations Test 21

$$2y.\dfrac{dy}{dx}=(2-x)^{2}-2x(2-x)$$
At $$(1,1)$$ we get 
$$2.\dfrac{dy}{dx}=1-2$$
Or 
$$\dfrac{dy}{dx}=\dfrac{-1}{2}$$
Now Equation of the tangent is 
$$(y-1)=\dfrac{-1}{2}(x-1)$$
Or 
$$2y-2=-x+1$$
Or 
$$x+2y=3$$
Or 
$$x=3-2y$$.
Substituting in the equation we get 
$$y^{2}=(3-2y)(2-3+2y)^{2}$$
Or 
$$y^{2}=(3-2y)(2y-1)^{2}$$
Or 
$$y^{2}=(3-2y)(4y^{2}-4y+1)$$
Or 
$$y^{2}=12y^{2}-12y+3-8y^{3}+8y^{2}-2y$$
$$y^{2}=-8y^{3}+20y^{2}-14y+3$$
Or 
$$8y^{3}-19y^{2}+14y-3=0$$
Solving the above cubic, we get
$$y=\dfrac{3}{8},\dfrac{3}{8}$$ and $$y=1$$
Considering $$y=\dfrac{3}{8}$$ we get 
$$x=3-2y$$
$$=3-\dfrac{3}{4}$$

$$=\dfrac{9}{4}$$
Hence the point is 
$$(\dfrac{9}{4},\dfrac{3}{8})$$

Let $$x=3\sqrt{3}\cos\theta$$
And 
$$y=\sin\theta$$
Hence, $$\dfrac{dy}{dx}$$
$$=-\dfrac{\cos\theta}{3\sqrt{3}\sin\theta}$$ ...(i)
Hence equation of the tangent will be 
$$y-\sin\theta=-\dfrac{\cos\theta}{3\sqrt{3}\sin\theta}(x-3\sqrt{3}.\cos\theta)$$
$$3\sqrt{3}\sin\theta.y-3\sqrt{3}\sin^{2}\theta=-x\cos\theta+3\sqrt{3}\cos^{2}\theta$$
$$3\sqrt{3}\sin\theta.y+x\cos\theta=3\sqrt{3}$$.
Hence the intercepts are 
$$cosec\theta$$ and $$3\sqrt{3}.\sec\theta$$.
Hence, $$k=cosec\theta+3\sqrt{3}.\sec\theta$$
Now $$\dfrac{dk}{d\theta}$$ $$=-cosec\theta.\cot\theta+3\sqrt{3}.\sec\theta.\tan\theta$$ $$=0$$
$$3\sqrt{3}.\sec\theta.\tan\theta=cosec\theta.\cot\theta$$
$$3\sqrt{3}.\tan^{2}\theta.\dfrac{1}{\cos\theta}=\dfrac{1}{\sin\theta}$$
$$3\sqrt{3}.\tan^{2}\theta=\dfrac{\cos\theta}{\sin\theta}$$
$$3\sqrt{3}.\tan^{3}\theta=1$$
$$\tan^{3}\theta=\dfrac{1}{3\sqrt{3}}$$
$$\tan\theta=\dfrac{1}{\sqrt{3}}$$
$$\theta=\tan^{-1}(\dfrac{1}{\sqrt{3}})$$
$$\theta=30^{0}$$ $$=\dfrac{\pi}{6}$$.

Given, $$3xy^{2}-2x^{2}y=1$$

$$f(x)=x^3+ax+b$$
$$f'(x)=3x^2+a$$
Given  gradient at $$x=a$$ and at $$x=b$$ are same
$$\Rightarrow 3a^2 +a=3b^2+a\Rightarrow b^2=a^2$$
But given $$a\neq b$$
$$\Rightarrow a+b=0 ..(1)$$
Hence $$ f(1)=1+a+b=1$$ using (1)

Let any point on the curve is $$\displaystyle \left ( \frac{c}{t},ct^{2} \right )$$
$$\displaystyle y=\frac{c^{3}}{x^{2}}\Rightarrow \frac{dy}{dx}=-\frac{2c^{3}}{x^{3}}=-\frac{2c^{3}}{\dfrac {c^{3}}{t^{3}}}$$
$$\displaystyle \frac{dy}{dx}=-2t^{3}$$
Equation of tangent is
$$\displaystyle y-ct^{2}=-2t^{3}\left ( x-\frac{c}{t} \right )$$
for x intercept
$$\displaystyle 0-ct^{2}=-2t^{3}\left ( a-\frac{c}{t} \right )\Rightarrow \frac{c}{2t}=a-\frac{c}{t}$$
$$\displaystyle a=\frac{3}{2}\frac{c}{t}$$
for y intercept
$$\displaystyle b-ct^{2}=-2t^{3}\left ( 0-\frac{c}{t} \right )$$
$$\displaystyle \Rightarrow b-ct^{2}=2t^{2}c  \:\:\Rightarrow b=3ct^{2}$$
$$\displaystyle a^{2}b=\frac{9}{4},\frac{c^{2}}{t^{2}}\times 3ct^{2}=\frac{27c^{3}}{4}$$

Given, $$\displaystyle y=ax^{4}+bx^{3}+cx+d$$
$$\Rightarrow y'=4ax^3+3bx^2+c$$
Using given conditions,
$$y(0)=1\Rightarrow d=1$$
$$y'(0)=0\Rightarrow c=0$$
$$y(-1)=0\Rightarrow a-b=-1 ..(1)$$
and $$y'(-1)=0\Rightarrow 4a-3b=0 ..(2)$$
Solving equation (1) and (2) we get, $$a=3,b=4$$
Hence the polynomial is,
$$y=3x^4+4x^3+1$$
$$y'=12x^2(1+x)$$
Now for negative gradient
$$y' < 0\Rightarrow 12x^2(1+x)< 0$$

We first determine the value of $$t$$ corresponding to the given values ofx and $$y$$. From $$t^{2}+3t-8= 2$$, we get $$t = 2, -5$$, and from $$2t^{2}-2t-5= 2$$ we get $$t = 2, -1$$. Hence to the given point there corresponds the value $$t = 2$$. Therefore, the slope of the tangent at $$\left ( 2,-1 \right )$$ is 

$$\displaystyle \left | y' \right |_{t=2}=\left | \dfrac{dy/dt}{dx/dt} \right |_{t=2}\:=\left | \dfrac{4t-2}{2t+3} \right |_{t=2}=\:\dfrac{6}{7}$$

$$3x+4y=c$$
Or 
$$3+4y'=0$$
Or 
$$y'=\dfrac{-3}{4}$$.
This is the slope of the tangent.
Now 
Differentiating the equation of the curve with respect to $$x$$, we get 
$$2x^{3}=1+y'$$
Now 
$$y'=\dfrac{-3}{4}$$
Hence
$$2x^{3}=\dfrac{1}{4}$$
Or 
$$x^{3}=\dfrac{1}{8}$$
Hence
$$x=0.5$$
Substituting in the equation of the curve we get 
$$\dfrac{1}{32}=\dfrac{1}{2}+y$$
Or 
$$y=\dfrac{-15}{32}$$.
Hence substituting the above points on the equation of the line, we get 
$$\dfrac{3}{2}+\dfrac{-15}{8}=c$$
Or 
$$c=\dfrac{-3}{8}$$
Hence there is only $$1$$ value of $$c$$.