Problem 35 - Analytical Geometry: 2D and 3D

Question

Prove that the polar of any point on \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with respect to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) touches \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 .\)

Step-by-Step Solution

Answer

Question: Prove that the polar of any point on the given ellipse with respect to the given hyperbola touches the ellipse. Answer: The polar of any point on the ellipse with respect to the hyperbola touches the ellipse because the point of tangency lies on the ellipse, as shown in the step-by-step solution provided.

Step 1: Find the equation of the polar with respect to the hyperbola

For a point P(x1, y1) on the ellipse, the equation of the polar with respect to the hyperbola is given by: \(\frac{xx_1}{a^2} - \frac{yy_1}{b^2}= 1\)

Step 2: Calculate the slope of the polar

Differentiating the polar equation with respect to x, we get: \(\frac{dx}{dy} = -\frac{y1 b^2}{x1 a^2}\)

Step 3: Find the point of tangency on the ellipse

Using the point-slope formula, we get the equation of line: \(y - y1 = -\frac{y1 b^2}{x1 a^2}(x - x1)\) Rearrange the equation to have it in the general form by multiplying both sides by \(x_1a^2\): \(y1 x1a^2 -y1^2 a^2 = -x y1 b^2 + x1^2 b^2\)

Step 4: Substitute the point P(x1, y1) from the ellipse equation

Since P(x1, y1) lies on the ellipse, we have: \(\frac{x1^2}{a^2} + \frac{y1^2}{b^2} = 1\) Rearrange the equation and substitute in the equation of line: \(y1^2 a^2 = x1^2 a^2 - x1^2 b^2\) Now substitute this equation into the line equation obtained in step 3: \(y1 x1a^2 - (x1^2 a^2 - x1^2 b^2) = -x y1 b^2 + x1^2 b^2\)

Step 5: Show that the point lies on the ellipse

Now, simplify the equation from step 4: \(x1^2 b^2 = x1^2 b^2 - x y1 b^2\) \(x1^2 b^2 = -xy1b^2 + x1^2 b^2\) Divide both sides by \(x1^2 b^2\) to get: \(1 = -\frac{xy_1}{x_1} + 1\) Since this equation is true, we can conclude that the point of tangency lies on the ellipse. Hence, the polar of any point on the ellipse with respect to the hyperbola touches the ellipse.

Key Concepts

Analytical Geometry and the Polar of a Point

Analytical geometry is a branch of mathematics that enables the study and description of geometric figures using algebra and the Cartesian coordinate system. In the context of conic sections like ellipses and hyperbolas, analytical geometry allows us to define and analyze these curves through their equations.

Consider the hyperbola \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\). Its polar for a given point \(P(x_1, y_1)\) is a line that can be considered a geometric locus of points from which tangents drawn to the hyperbola pass through \(P\). By analyzing the equation of the polar line and applying the point-slope form, we can draw connections between the different conic sections and identify points of tangency through algebraic manipulation.

Using differentiation and substitution as shown in the original exercise, we transform the geometry problem into an algebraic one. Simplifying the equations reveals the point of tangency and confirms the property that the polar of a point on the ellipse indeed touches the ellipse, as tangency implies that the polar intersects the ellipse at exactly one point.

Tangency of Polar

Tangency of polar is a concept that deals with how a polar line is related to a conic section such as a hyperbola or an ellipse. In simple terms, a polar line of a point with respect to a conic will be tangent to the conjoined conic section if it intersects at precisely one point.

For the given exercise, we know that the polar line is determined based on a point \(P\) on the ellipse. By differentiating the equation of the polar and using algebra, we are able to show that the line only touches the ellipse at \(P\). This property is fundamental in the study of conic sections and has implications in other fields such as optics and mechanical engineering, where understanding the tangency is crucial for lens design and the analysis of dynamic systems respectively.

The step-by-step solution outlined previously is a practical implementation of this principle. It confirms that the polar line, derived from the point on the ellipse, has a slope equal to the tangent of the curve at that point on the ellipse, thus satisfying the tangency condition.

Hyperbola and Ellipse Properties

Hyperbolas and ellipses are two types of conic sections defined by their distinct geometric properties and their standard equations in the Cartesian plane. Hyperbolas open in opposite directions and have equations of the form \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\), whereas ellipses are closed curves with equations like \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\).

One interesting property arising from these equations is their reflection and duality principles. If a point lies on an ellipse, its polar with respect to the conjugate hyperbola will be a tangent to the ellipse. Conversely, if a line is a tangent to a hyperbola, its pole with respect to the hyperbola will lie on the conjugate ellipse.

Through the exercise solution, we see a demonstration of how the geometry of these conic sections translates into algebraic relations. By proving that the polar of a point on an ellipse touches the ellipse, we're essentially verifying those reflective and dualistic properties, showing a deep interconnectedness between hyperbolas and ellipses that extends beyond their simple definitions.