Tangents and it...

At the point $$P(a,a'')$$ on the graph of $$y=x^n$$, $$(n \epsilon N)$$, in the first quadrant , a normal is drawn. The normal intersects the $$y$$-axis at the point $$(0,b)$$. If $$\lim_{a\rightarrow 0}b=\displaystyle \frac{1}{2}$$, then n equals

Let $$f(x)$$=$$\begin{cases} -x^2, {for  \   x<0} \\x^2+8,  {for \   x\geq 0}  \end{cases}$$. Then $$x$$-intercept of the line, that is, the tangent to the graph of $$f(x)$$ in both the intervals of its domain, is

A function $$y=f(x)$$ has a second-order derivative $$f''(x)=6(x-1)$$. If its graph passes through the point $$(2,1)$$ and at the point tangent to the graph is $$y=3x-5$$, then the value of $$f(0)$$ is 

The tangent of the acute angle between the curves $$y=|x^2-1| $$ and $$y=\sqrt {7-x^2}$$ at their points of intersection is

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

The abscissas of points $$P$$ and $$Q$$ on the curve $$y=e^x+e^{-x}$$ such that tangents at $$P$$ and $$Q$$ make $$60^{\circ}$$ with the $$x$$-axis are

The point on the curve $$y^{2} = x ,$$ the tangent at which makes an angle of $$45^{0}$$ with positive direction of $$x -$$ axis will be given by

If the curve represented parametrically by the equations $$x=2 \ln\cot t+1$$ and $$y=\tan t+ \cot t$$

A curve is represented by the equations, $$\displaystyle x = \sec^2t$$ and $$y =\cot t$$, where $$t$$ is a parameter. If the tangent at the point $$P$$ on the curve where $$\displaystyle t = \dfrac{\pi}{4}$$ meets the curve again at the point $$Q$$ then $$|PQ|$$ is equal to

The x-intercept of the tangent at any arbitrary point of the curve $$\displaystyle \frac {a}{x^2} + \frac {b}{y^2} = 1$$ is proportional to: