Tangents and it...

The angle made by the tangent at any point on the curve $$x=a(t+\sin { t } \cos { t } ),y=a{ (1+\sin { t } ) }^{ 2 }$$ with x-axis is

If tangent at any point on the curve $${ y }^{ 2 }=1+{ x }^{ 2 }\ makes\ an\ angle\ \theta $$ with positive direction of the x-axis then

The equation of the normal to the curve $$y=(1+x)^{ y }+\sin { ^{ -1 }(\sin ^{ 2 }{ x)\ at\ x=0\ is }  } $$.

If the line $$ax+y=c$$, touches both the curves $${x}^{2}+{y}^{2}=1$$ and $${y}^{2}=4\sqrt{2}x$$, then $$\left| c \right| $$ is equal to:

The equation of the normal to the curve$$y=\left( 1+x \right) ^{ y }+{ sin }^{ -1 }\left( { sin }^{ 2 }x \right) at\quad x=0$$ is

The sum of the length of sub tangent of sub tangent and tangent to the curve
$$x=c\left[ 2cos\theta -log\left( cos\quad ec\theta +cot\theta  \right)  \right] ,y=csin2\theta \quad at\quad \theta =\frac { \pi  }{ 3 } is$$

Length of the normal to the curve at any point on the curve $$y=\dfrac { a\left( { e }^{ x/a }+{ e }^{ -x/a } \right)  }{ 2 } $$ varies as 

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

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

At what points the slope of the tangent to the curve $$x^2+y^2-2x-3=0$$ is zero?