Tangents and its Equations Test 50

Differentiating the equation of the curve with respect to x, we get 
$$y+xy'=0$$
Or 
$$y'=\dfrac{-y}{x}$$

$$=\dfrac{-4}{x^{2}}$$

$$y'_{x=1}=-4$$
Hence slope of normal is 
$$=\dfrac{1}{4}$$.
Hence equation of normal will be 
$$y-4=\dfrac{1}{4}(x-1)$$
Or 
$$4y-16=x-1$$
Or 
$$4y-x=15$$ ...(i)
$$xy=4$$ ...(ii)
Solving the above two equations
$$4y-\dfrac{4}{y}=15$$
Or 
$$4y^{2}-4=15y$$
$$4y^{2}-15y-4=0$$
$$y=4$$ and $$y=\dfrac{-1}{4}$$
Hence
$$y=\dfrac{-1}{4}$$.
Thus $$x=\dfrac{4}{y}=-16$$.
Hence the required point is 
$$(-16,\dfrac{-1}{4})$$.

We require the point where $$\dfrac{dx}{dy}=0$$
Or 
$$3y^{2}.dy+6x.dx=12dy$$
Or 
$$dy(3y^{2}-12)=-6xdx$$
Or 
$$\dfrac{-dx}{dy}=\dfrac{3y^{2}-12}{6}$$
$$=0$$
Hence
$$3y^{2}-12=0$$
$$y^{2}-4=0$$
$$y=\pm 2$$
Now $$y=-2$$ does not satisfy the equation the curve.
Substituting in the equation of the curve, y=2,
$$8+3x^{2}=24$$
$$3x^{2}=16$$
$$x=\pm\dfrac{4}{\sqrt{3}}$$.

Given, $$y=x^4$$

The given equation $$\displaystyle x^{2}=y-1 $$ represents a parabola with vertex at $$(0,1)$$
Let the co-ordinates of any point p on the curve $$\displaystyle y=1+x^{2}be\left ( h,1+h^{2} \right ).$$
 Now $$\displaystyle \dfrac{dy}{dx}=2x=2h $$ Tangent at P is $$\displaystyle y-\left ( 1+h^{2} \right )=2h\left ( x-h \right )$$ or $$\displaystyle y-2xh=1-h^{2}$$ 
Substituting $$x=0$$ , we get $$\displaystyle y=\left ( 1-h^{2} \right )OL $$ 
Substituting $$x=1$$, we get $$\displaystyle y=1+2h-h^{2}$$ 
$$\displaystyle \therefore M is \left ( 1,1+2h-h^{2} \right ),N\left ( 1,0 \right )$$
A=area of trapezium $$\displaystyle =\dfrac{1}{2}\left ( OL+MN \right ).ON$$ $$\displaystyle =\dfrac{1}{2}\left [ 1-h^{2}+12h-h^{2} \right ].1$$ 
$$\displaystyle A=(1+h-h^{2})$$ 
$$\displaystyle \therefore \dfrac{dA}{dh}=1-2h=0$$
$$ \therefore h=\dfrac{1}{2}$$ and $$\displaystyle \dfrac{d^{2}A}{dh^{2}}=-2=-ive $$ Hence max. 
$$\displaystyle \therefore $$ Point P is $$\displaystyle \left ( \dfrac{1}{2}.\dfrac{5}{4} \right ).$$ 

Ans: B

$$x=a\left( \theta +sin\theta  \right) $$
$$\cfrac { dx }{ d\theta  } =a\left( 1+cos\theta  \right) $$
$$y=a\left( 1-cos\theta  \right) $$
$$\cfrac { dy }{ d\theta  } =asin\theta $$

Then, $$\cfrac { dy }{ dx } =\cfrac { asin\theta  }{ a\left( 1+cos\theta  \right)  } $$

$$ke^{ x }-x=0$$ will have only one solution when $$g(x)=x$$ is tangent to $$f(x)=ke^{x}$$
let $$P(a,b)$$ be a point on $$f(x)$$ and it also satisfy $$g(x)$$
Therefore, $$ke^{ a }=a$$       ...(1)
slope of $$g(x)=$$slope of $$f(x)$$ at point P
$$\Rightarrow 1=\dfrac { dy }{ dx } $$at point P$$=k{ e }^{ a }$$
From (1), we get $$a=1$$
Therefore, $$k=1/e$$

Ans: A