Tangents and it...

The equation of the normal to the curve $${ y }^{ 4 }=a{ x }^{ 3 }$$ at $$\left( a,a \right) $$ is

If tangent to the curve $$\displaystyle x={ at }^{ 2 },y=2at$$ is perpendicular to $$x$$-axis, then its point of contact is:

The slope of the normal to the curve $$y = 2x^2+ 3 \sin x$$ at $$x = 0$$ is. 

The slope of the tangent to the curve $$y=\displaystyle\int_{0}^{x}\dfrac{dt}{1+t^3}$$ at the point where x=1 is 

The equation of the tangent to the curve $$x=\sqrt{t}, y=t-\dfrac{1}{\sqrt{t}}$$ at $$t=4$$ is:

The chord of the curve $$y = x^{2} + 2ax + b$$, joining the points where $$x = \alpha$$ and $$x = \beta$$, is parallel to the tangent to the curve at abscissa $$x =$$

Consider the curve $$y = e^{2x}$$.What is the slope of the tangent to the curve at (0, 1) ?

If normal is drawn to $${ y }^{ 2 }=12x$$ making an angle $${45}^{o}$$ with the axis then the foot of the normal is

The equation of the normal to the curve $$y=-\sqrt { x } +2$$ at the point of its intersection with the bisector of the first quadrant is

Equation of normal drawn to the graph of the function defined as $$f\left( x \right) =\dfrac { \sin { { x }^{ 2 } }  }{ x } ,x\neq 0$$ and $$f\left( 0 \right) =0$$ at the origin is