Find the point on the curve $y = \sqrt{4x - 3} - 1$ where the tangent has a slope of $\frac{2}{3}$.

The coordinates of the point on the curve $y=x^2-3x+2$ where the tangent is perpendicular to the straight line $y=x$ are

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

Find the equation of the normal lines to the curve $3x^{2}-y^{2}=8$ which are parallel to the line $x+3y=4$.

If the point $P(x_1, y_1)$ lying on the curve $y = x^2 - x + 1$ is the closest point to the line $y = x - 3$,then the perpendicular distance from $P$ to the line $3x + 4y - 2 = 0$ is

For $h, k \in N$,let $P(h, k)$ be the point of intersection of the curves $x^2 y - x^3 = 8$ and $y^3 - x y^2 = 32$. If $\theta$ is the acute angle between these two curves at $P$,then $\tan \theta =$