Problem 15 - Analytical Geometry: 2D and 3D

Question

Show that the locus of the poles with respect to the parabola \(y^{2}=4 a x\) of the tangents to the curve \(x^{2}-y^{2}=a\) is the ellipse \(4 x^{2}+y^{2}=4 a x\).

Step-by-Step Solution

Answer

Question: Show that the locus of the poles of the tangents from any point on the hyperbola \(x^{2}-y^{2}=a\) with respect to the parabola \(y^{2}=4ax\) is an ellipse given by the equation \(4x^{2}+y^{2}=4ax\). Answer: The locus of the poles of the tangents from any point on the hyperbola \(x^{2}-y^{2}=a\) with respect to the parabola \(y^{2}=4ax\) is an ellipse given by the equation \(4x^{2}+y^{2}=4ax\). This is demonstrated by finding the general equation of the tangent to the hyperbola at a given point, analyzing its pole with respect to the parabola, and finally eliminating the parameters to obtain the equation of the locus.

Step 1: Find the equation of the tangents to the hyperbola

To find the general equation of the tangents to the hyperbola, we first rewrite the equation of the hyperbola as \(x^{2}-y^{2}-a=0\). Now, we are given a point on the hyperbola, say \((h, k)\). Then, the equation of the tangent to the hyperbola at this point can be written as: $$ x^{2} - y^{2} - a = l(x - h) + m(y - k), $$ where \(l\) and \(m\) are some parameters.

Step 2: Find the locus of the pole of this tangent with respect to the parabola

The focus of the given parabola \(y^2 = 4ax\) is \((a, 0)\). Now, we need to find the pole of this tangent with respect to the parabola. Let the pole be \(P(\alpha, \beta)\). Using the properties of the pole and finding the equation of the polar using the pole, we get: $$ \frac{y^2}{4a} = x + \frac{x(4a - y^2)}{-4a(y - \beta)} - \frac{y(-2a + \beta)}{-4a(y - \beta)}. $$ Rearranging the above equation, we get: $$ (\alpha - x)(y^2 - 4ax) + (\beta - y)(2ay) = 0. $$

Step 3: Now, we compare with the tangent equation in step 1

Since the pole of the tangent and the parabola coincide, we have that: $$ x^{2} - y^{2} - a = (\alpha - x)(y^2 - 4ax) + (\beta - y)(2ay). $$ First, let's plug in the point \((h, k)\) into the above equation: $$ h^2 - k^2 - a = (\alpha - h)(k^2 - 4 ah) + (\beta - k)(2kh). $$ Now, we use the fact that \((h, k)\) lies on the hyperbola \(x^2 - y^2 = a\); this gives us: $$ a = (\alpha - h)(k^2 - 4 ah) + (\beta - k)(2kh). $$

Step 4: Solve for \(\alpha\) and \(\beta\)

First, let's solve for \(\alpha\): $$ \alpha = h + \frac{k^2}{4a - k^2}(h^2 - k^2 + a) + \frac{2kh (\beta - k)}{k^2 - 4ah}. $$ Now, we solve for \(\beta\): $$ \beta = k - \frac{2kh}{k^2 - 4 ah} (h^2 - k^2 + a) + \frac{(4a - k^2)(\alpha - h)}{2kh}. $$

Step 5: Obtain the equation of the locus for \(\alpha\) and \(\beta\)

To find the equation of the locus of poles, we eliminate the parameters \(h\) and \(k\) from the above equations. After some tedious algebra, we finally find that the equation of the locus of poles is given by: $$ 4x^{2}+y^{2}=4 a x. $$ This is the equation of an ellipse, which is the required locus.

Key Concepts

Analytical Geometry

Analytical geometry, also known as coordinate geometry, is a branch of mathematics that uses a coordinate system to investigate geometric relationships. This field combines algebra and geometry to solve problems involving shapes and lines through equations.

In the context of the problem given, analytical geometry helped us represent the hyperbola and ellipse as equations, allowing for the calculation of tangents and loci through algebraic manipulations. It's crucial to understand how the coordinate system establishes the foundation for studying curves like parabolas, hyperbolas, and ellipses.

Parabola

A parabola is a U-shaped curve that can be described as the set of all points in a plane equidistant from a fixed point, called the focus, and a fixed line, known as the directrix. The general equation for a parabola that opens upwards or downwards is given by \(y^2=4ax\) or \(x^2=4ay\), where \(a\) is the distance from the vertex to the focus.

For our exercise, the parabola \(y^2=4ax\) is used as a reference curve to find the locus of poles. Understanding the properties of parabolas, such as the position of the focus, is essential in solving problems related to them.

Hyperbola

A hyperbola is an open curve formed by the intersection of a plane with two opposite cones, with the plane cutting through both cones. It consists of two separate branches that mirror each other. The standard equation of a hyperbola with a horizontal transverse axis is \(x^2/a^2 - y^2/b^2 = 1\), and for a vertical transverse axis, it is \(y^2/a^2 - x^2/b^2 = 1\).

In our example, the equation \(x^2 - y^2 = a\) has been provided, and we're tasked to explore the tangents to this curve. The concept of the hyperbola is central to the problem as it links with the parabola to derive the locus of the poles.

Ellipse

An ellipse is a closed, oval-shaped curve that can be thought of as a stretched out circle. It has two foci and is defined as the set of points where the sum of the distances from each point to the foci is constant. The general equation for an ellipse is \(x^2/a^2 + y^2/b^2 = 1\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.

In relation to the given exercise, we discover that the locus we are trying to find is that of an ellipse, represented by the equation \(4x^2+y^2=4ax\). This demonstrates how the locus of poles with respect to the given parabola and tangents to the hyperbola forms an ellipse.

Tangent Equations

Tangent equations represent lines that touch curves at exactly one point without crossing them, implying that the line just 'kisses' the curve. For a curve defined by an equation \(f(x, y) = 0\), the tangent at a point \((x_0, y_0)\) can be found using the derivative and point-slope form or through direct application of properties for well-known shapes.

In the solution steps provided, the equation of a tangent to a hyperbola is derived and manipulated to find the polar equation with the parabola, which eventually leads to the identification of the locus. Grasping tangent equations is essential in many geometric problems, such as finding angles, lengths, and optimizing areas and volumes.