Find the slope of the tangents to the curve $y = (x + 1)(x - 3)$ at the points where it meets the $x$-axis.

If the length of the subnormal is equal to the length of the subtangent at any point on the curve $y = f(x)$ and the tangent at $(3, 4)$ to $y = f(x)$ meets the positive coordinate axes at $A$ and $B$,then the area of $\Delta OAB$,where $O$ is the origin,is

Let $f(x) = \begin{cases} -x^2 & \text{for } x < 0 \\ x^2 + 8 & \text{for } x \ge 0 \end{cases}$. Then the $x$-intercept of the line that is tangent to the graph of $f(x)$ is

The length of the perpendicular drawn from the origin on the normal to the curve $x^2+2xy-3y^2=0$ at the point $(2,2)$ is

The point on the curve $y=x^2+4x+3$ which is closest to the line $y=3x+2$ is

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