A focused parabola

I came up with a solution that doesn't require knowledge of the formula(s) for the focus:

Let $A$ be the focus, $O$ be the origin and $B$ the point $(x_0, y_0)$ . Since the minimum point is the origin, the equation must be of the form $y = ax^2$ .

Now each point on the parabola is equidistant from the focus and directrix. Since $O$ is a distance of $1$ from the focus, therefore the directrix must be $y = -1$ .

It is also known that $B$ is $101$ units away from the focus. Hence it is also $101$ units from the directrix (i.e. the two orange lines in the diagram are of equal length). Thus $y_0 = -1 + 101 = 100$ .

Let $C$ be the point $(x_0, 1)$ as in the diagram. Then, in the triangle $ABC$ we have $BC = 99, AB = 101$ and $\angle ACB = 90^{\circ}$ . Therefore, by the Pythagorean Theorem, $AC = \sqrt{101^2 - 99^2} = 20$ .

Thus $x_0 = 20$ and $y_0 = 100$ . Since $y_0 = ax_0^2$ , this gives us $a = \dfrac{1}{4}$ . Hence the parabola is $y = \dfrac{1}{4} x^2$ . The derivative of the parabola is $\dfrac{x}{2}$ , so when $x = 20$ the slope of the tangent is $\boxed{10}$ .

The parabola is described by $y = x^{2}/4$ , which can easily be derived.

There is a property of parabolas: a parabola can be defined as the locus of points which are equidistant from a focus as well as a directrix. Using this, we then know that $y_{0} = 100$ , since the directrix is $y_{\text{dir}} = -1$ . Hence, we get that $x_{0} = 20$ .

Now, to find the tangent, we simply differentiate:

$\left. \frac{dy}{dx} \right|_{x=x_{0}} = x_{0}/2 = \boxed{10}$

For a parabola of the form $y = a(x - h)^2 + k,$ the critical point is located at the co-ordinates $(h, k)$ , and the focus is located at $(h, k + \frac{1}{4a})$ .

Given that the minimum of the parabola is at $(0, 1)$ , and the minimum — which is a type of critical point — is located at $(0, 0)$ . Therefore, $h = k = 0$ So $\frac{1}{4a} = 1$ $a = \frac{1}{4}$

Therefore the equation of the parabola is $y = \frac{1}{4}x^2$ . Differentiating, we get $\frac{\mathrm d y}{\mathrm d x} = \frac{1}{4}(2)x^{2-1}= \frac{1}{2}x$ .
We'll need this to find the gradient described in the problem.

A line of length $101$ units is drawn from the focus to the point $(x_0, y_0)$ on the parabola. Given that $(x_0, y_0)$ lies in the first quadrant, find the gradient of the tangent at $x = x_0$ .

The first thing we want to find is (effectively) the point of intersection between the parabola, and the circle with centre $(0, 1)$ and radius $101$ . So, we have $(x - 0)^2 + (y - 1)^2 = 101^2.$ Substituting in for $y$ , $x^2 +\bigg (\frac{x^2}{4}\bigg)^2 = 101^2$ $x^2 +\bigg [\frac{x^4}{16} - 2\bigg(\frac{x^2}{4}\bigg) + 1\bigg] = 10201$ $\frac{x^2}{2} + \frac{x^4}{16} = 10200$ $x^4 + 8x^2 = 163200$ Let $z = x^2$ . So $z^2 + 8z = 163200$ Therefore, $(z+4)^2 = (4)^2 + 163200$ $(z+4)^2 = 163216$ $z+4 = \pm\sqrt{163216} = \pm404$ $z = -4\pm404$ Ignore negative solution, as $x^2$ is real. Therfore $z = 400$ Substituting in for $z$ , $x^2 = 400$ $x = \pm\sqrt{400} = \pm20$ However, $x_0$ is in the first quadrant, making it positive. So $x_0 = 20$

All that is left to do is to find the tangent at $x = 20$ . Plugging in this value for the formula for the derivative, we acquire $\frac{1}{2}(20) = \boxed{10},$

yielding our final answer.