Tangents and it...

The points on the graph $$ y = x^3 - 3x $$ at which the tangent is parallel to $$x$$-axis are :

If the tangent at $$(1,1)$$ on $${ y }^{ 2 }=x{ (2-x) }^{ 2 }$$ meets the curve again at $$P$$, then $$P$$ is

The tangent to the curve $$y=a{ x }^{ 2 }+bx$$ at $$\left( 2,-8 \right) $$ is parallel to $$X$$-axis. Then,

If the straight line $$ y -2x +1=0$$ is the tangent to the curve $$xy+ax+by=0$$ at $$x=1, $$ then the values of $$a$$ and $$b$$ are respectively :

The slope of the normal to the curve $$x = t^{2} + 3t - 8$$ and $$y = 2t^{2} - 2t - 5$$ at the point $$(2, -1)$$ is

If the angle between the curves $$ y = 2^x $$ and $$ y=3^x $$ is $$ \alpha, $$ then the value of $$ \tan \alpha $$ is equal to :

The slope of the normal to the curve $$ y^3 - xy-8=0$$ at the point $$(0,2)$$ is equal to :

The equation of the tangent to the curve $$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1$$ at the point $$(x_{1}, y_{1})$$ is $$\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = k$$. Then, the value of $$k$$ is

The slope of the normal to the curve $$y = x^2 - \dfrac{1}{x^2}$$ at $$(-1, 0) $$ is 

The point on the curve $$y = 5 + x - x^{2}$$ at which the normal makes equal intercepts is