Given the equation x²/9 + y²/4 = 1 find the set of all points from which the angle of view^* of this ellipse is a right angle. What is the significance of this set of points?

Given the equation x²/9 - y²/4 = 1 find the set of all points from which the angle of view^* of this hyperbola is a right angle. What is the significance of this set of points?

* i.e. displaying a right angle between the two tangents.

I agree with Bractal’s algebraic results and use his notation, with s = +1 or -1:

A line through P(X, Y) with gradient m has equation y = Y + m(x – X)

and any points of intersection with the conic section  x²/a² + sy²/b² = 1 will

satisfy the equation   x²/a² + s[Y + m(x – X)]²/b² = 1        which can be

written as a quadratic in x as follows:

x²(b² + sm²a²) + x[2sma²(Y – mX)] + sa²(Y – mX)² – a²b²= 0

For the line to be a tangent to the conic this equation has equal roots for x.

Thus   [2sma²(Y – mX)]² = 4((b² + sm²a²)[ sa²(Y – mX)² – a²b²]

which simplifies to:          m²s(X² – a²) – m(2sXY) + sY² – b² = 0

For the two possible tangents from P to be perpendicular, the product of the roots of this quadratic in m must be -1.

Thus     sY² – b² = -s(X² – a²), giving        X² + Y² = a² + sb²           (1)

For s = 1 this is a circle, centre (0, 0) and radius sqrt(a² + b²)

For s = -1 this is, potentially, a circle, centre (0, 0), radius sqrt(a² – b²), but

the original conic is a hyperbola with  two distinct branches and with

asymptotes y = (b/a)x and y = -(b/a)x dividing the plane into four regions.

This means that two tangents to the same branch can only meet at points within the region occupied by that branch.

Case (i)  If a < b then (1) has no solution and no possible points exist.

Case (ii)  If a >= b then possible points lie on the two arcs of the circle (1)

which lie in the regions |Y| < (b/a)|X| containing the branches.

However, it seems to me that, while it makes sense to use tangents to define the limits of what can be seen of a closed curve, in the case of a hyperbola viewed from a point ‘outside’ the asymptotes of a particular branch, the ‘edge’ of that branch would be in the direction of the point at infinity on the asymptote; in other words, in the direction parallel to the asymptote.
This means that the two circular arcs found above can be extended tangentially

into the regions |Y| > (b/a)|X| to produce four rays drawn on the parallelogram

bY = +/- [aX +/- sqrt(a⁴ – b⁴)] which meet in pairs at (0, +/- sqrt(a⁴ – b⁴)/b)

In our case (a =3, b = 2) we have arcs on X² + Y² = sqrt(5) with rays on

2Y = +/- [3X +/- sqrt(65)], which meet at (0, +/- sqrt(65)/2) from which

points both  branches of the hyperbola present a right angle.

Posted by Harry on 2011-11-25 23:17:28