Tangents and it...

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:

The normal to the curve $$x=a\left( \cos { \theta  } +\theta \sin { \theta  }  \right) $$, $$y=a\left( \sin { \theta  } -\theta \cos { \theta  }  \right) $$ at any point $$\theta$$ is such that

The slope of the curve $$y=\sin { x } +\cos ^{ 2 }{ x }$$ is zero at a point , whose x-coordinate can be ?

The tangent to the circle $$x^2+y^2=5$$ at the point $$(1,-2)$$  also touches the circle $$x^2+y^2-8x+6y+20=0$$ at 

If for a curve represented parametrically by $$x={ sec }^{ 2 }t,\quad y=cot\quad t\quad $$ , the tangent  at a point $$P(t=\frac { \pi  }{ 4 } )$$ meets the curve again at the point Q, then $$\begin{vmatrix} PQ \end{vmatrix}$$is equal to 

If the subnormal to the curve $${ x }^{ 2 }.{ y }^{ n }={ a }^{ 2 }$$ is a constant then n=

If $$\theta$$ is angle of intersection between $$y=10-x^{2}$$ and $$y=4+x^{2}$$ then $$|\tan \theta|$$ is-

$$y = 6\tan \,x\left( {{e^x} - x - 1} \right) - 3{x^3} - {x^4} - \frac{5}{4}{x^5},\,$$ if $${n^{th}}$$ derivative at x=0 is non zero then least value of n is

The curve $$y-{e}^{xy}+x=0$$ has a vertical tangent at the point

The slope of the tangent to the curve at a point (x, y) on it is proportional to (x-2). If the slope of the tangent to the curve at  (10,-9) on it -3. The equation of the curve is