Tangents and it...

Let $$C$$ be a curve given by $$y(x) = 1 + \sqrt {4x - 3}, x > \dfrac {3}{4}$$. If $$P$$ is a point on $$C$$, such that the tangent at $$P$$ has slope $$\dfrac {2}{3}$$, then a point through which the normal at $$P$$ passes, is:

What is the $$x$$-coordinate of the point on the curve $$f(x) = \sqrt {x}(7x - 6)$$, where the tangent is parallel to $$x$$-axis?

Tangents are drawn from the origin to the curve $$y=\cos { x } $$. Their points of contact lie on

Determine the equation of tangent at vertex of the parabola $$\displaystyle (x+4)^{2}=-4(y-2)$$.

Normal to the parabola $$\displaystyle y^{2}=4ax$$ where $$m$$ is the slope of the normal is

Tangent to parabola $$\displaystyle y^{2}=4x+5$$ which is parallel to $$y=2x+7$$

The slope of tangent to the curve $$y=\int_{0}^{x}\displaystyle \frac{dx}{1+x^{3}}$$ at the point where $$x=1$$ is

The values of '$$a$$' for which $$y=x^{2}+ ax+25$$ touches $$x$$-axis are

The equation of the normal to the curve $$\displaystyle 2y=3-x^{2}$$ at the point $$(1,1)$$

The equation of the normal to the curve $$\displaystyle x^{3}+y^{3}=6xy$$  at the point  $$(3,3).$$