Tangents and it...

If $$\left | f(x_{1})-f(x_{2}) \right |< (x_{1}-x_{2})^{2}$$ for all $$x_{1}$$ $$x_{2}$$ $$\in $$ R. Find the equation of tangent to the curve y = f(x) at the point (1, 2). 

For the curve $$y=3  \sin \theta  \cos  \theta,  x= e^{\theta} \sin \theta,  0  \leq \theta  \leq  \pi$$, the tangent is parallel to x-axis when $$\theta$$ is :

If the tangent to the curve $$y=\cfrac { x }{ { x }^{ 2 }-3 } ,x\in R,\left( x\neq \pm \sqrt { 3 }  \right) $$, at a point $$\left( \alpha ,\beta  \right) \neq \left( 0,0 \right) $$ on it is parallel to the line $$2x+6y-11=0$$, then:

The tangent to the curve $$y  =x^2 - 5x + 5$$, parallel to the line $$2y = 4x + 1$$, also passes through the point

The tangent at the point $$(2, -2)$$ to the curve, $$x^2y^2-2x=4(1-y)$$ does not pass through the point.

The normal to the curve $$y(x-2)(x-3)=x+6$$ at the point where the curve intersects the y-axis passes through the point.

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

If the tangent at $$(1, 7)$$ to the curve $$x^{2} = y - 6$$ touches the circle $$x^{2} + y^{2} + 16x + 12y + c = 0$$ then the value of $$c$$ is

If the tangent to the conic, $$y - 6 = x^2$$ at (2, 10) touches the circle, $$x^2 + y^2 + 8x - 2y = k$$ (for some fixed k) at a point $$(\alpha, \beta)$$; then $$(\alpha, \beta)$$ is;

The intercepts on $$x$$-axis made by tangents to the curve, $$y=\int_{0}^{x}|t|$$ dt, $$x\in R$$, which are parallel to the line $$y=2x$$, are equal to