The tangent drawn at the point $(0, 1)$ on the curve $y = e^{2x}$ meets the $x-$axis at the point:

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

If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x=12(t+\sin t \cos t)$,$y =12(1+\sin t )^{2}$,$0 < t < \frac{\pi}{2}$,with the positive $x$-axis is $\frac{\pi}{3}$,then $y_{0}$ is equal to

Let $m$ be the slope of the normal $L$ drawn at $(1, 2)$ to the curve $x = t^2 - 7t + 7, y = t^2 - 4t - 10$ and $ax + by + c = 0$ be the equation of the normal $L$. If the $G$.$C$.$D$. of $(a, b, c)$ is $1$, then $m(a + b + c) =$

If the line $x + By + C = 0$ is the normal to the curve given by $x = a \sin^3 t$ and $y = b \cos^3 t$ (where $a, b \neq 0$) at the point $t = \frac{\pi}{2}$, then $B - C =$

The lines tangent to the curves $y^3 - x^2y + 5y - 2x = 0$ and $x^4 - x^3y^2 + 5x + 2y = 0$ at the origin intersect at an angle $\theta$ equal to