Application of ...

The slope of the tangent to the curve $${r^2} = {a^2}\cos 2\theta$$, where $$x = r\cos \theta ,y = r\sin \theta $$, at the point $$\theta=\frac{\pi}{6}$$ is

The line $$\dfrac{x}{a}+\dfrac{y}{b}=1$$ touches the curve $$y=be^{-x/a}$$ at the point.

if $$m$$ is the slope of a tangent to the curve $$e^{y}=1+x^{2}$$, then  $$m$$ belongs to the interval

IF $$f(x)=\dfrac{x^2}{2-2cos x} ; \ g(x)=\dfrac{x^2}{6x-6sin x}$$ where $$0< \times  < 1$$, then 

A tangent drawn to the curve $$y = f\left( x \right)$$ at $$P\left( {x,y} \right)$$
cuts the x and y axes at A and B, respectively, such that $$AP:PB = 1:3$$. If $$f\left( 1 \right) = 1$$ then the curve passes through $$\left( {k,\frac{1}{8}} \right)$$ where $$k$$ is

On the curve $${x}^{3} = 12y$$ , then the interval at which the abscissa changes at a faster rate than the ordinate ?

Let $$f$$ and $$g$$ be differentiable function satisfying $$g'(a)=2,g(a)=b$$ and $$fog=I$$ (identity function). Then $$f'(b)$$ is equal to 

The point on the curve $$y = b e^{\dfrac {-x}{a}}$$ at which the tangent drawn is $$\dfrac {x}{a} + \dfrac {y}{b} = 1$$ is

If $$V$$ is the set of points on the curve $$y^{3} - 3xy +2 = 0$$ where the tangent is vertical then $$V =$$.

Paraboals $$(y-\alpha )^{2}=4a(x-\beta )and (y-\alpha )^{2}=4a'(x-\beta ')$$ will have a common normal (other than the normal passing through vertex ofparabola)if: