Focus with a hyperbola or circle?

Geometry Level 4

(Pre-requisite: focus-directrix description and/or eccentricity of conics)

A hyperbola following the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is plotted on the Cartesian plane. A tangent to the graph and the normal of the graph at the same point is drawn. The tangent has a $y$ -intercept of $\alpha$ and the normal has a $y$ -intercept of $\beta$ .

Let $\alpha = -4$ and $\beta = 9$

Find the $x$ -coordinate of the focus $c$ of the hyperbola, given that $c$ lies on the positive $x$ -axis.

(Hint: read the title)

The answer is 6.

1 solution

Lucas Tan
Dec 18, 2018

The $y$ -intercepts of the tangent and the normal and the 2 foci of the hyperbola all lie on a circle. My workings on this property can be found here: https://www.desmos.com/calculator/hzbyh9skzo

I didn't bother labelling and commenting my working, so you might prefer to prove it yourself. In the link, you'll need the understand polar equations to comprehend the moving graphs/tangent.

Once you have proven that the foci and the $y$ -intercepts all lie on the circle, note that the centre of the circle must lie on the $y$ -axis, since the centre of the hyperbola is at the origin. Finding the centre of the circle and its radius in terms of $\alpha$ and $\beta$ , you can obtain the equation of circle to be:

$x^2+\left(y-\frac{\left(\alpha+\beta\right)}{2}\right)^2=\frac{\left(\alpha-\beta\right)^2}{4}$

To find foci, or $x$ -intercept of circle, sub $y=0$ .

$\therefore x = \sqrt{-\alpha \beta}$ for focus $c$ on the positive $x$ -axis.

Hence $x$ -coordinates of $c = \sqrt{-(-4)(9)} = 6$

Focus with a hyperbola or circle?

The answer is 6.

1 solution

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