Question

Show that the tangent to the hyperbola $$r=\frac{m}{1-\sqrt{3} \cos \phi}$$ at the extremity of the latus rectum makes an angle of \(60^{\circ}\) with the transverse axis.

Short answer

In summary, given the polar equation of the hyperbola \(r = \frac{m}{1 - \sqrt{3}\cos\phi}\), we found the Cartesian equation as \(x + \sqrt{3}y = m\), and the extremity of the latus rectum as \(\left(\frac{m}{\sqrt{3}}, \frac{m}{2}\right)\). The tangent at this point has a slope of \(\frac{3}{2\sqrt{3}-3}\), and it makes an angle of \(60^{\circ}\) with the transverse axis.

Step-by-step solution

Find the Cartesian equation of the hyperbola

Given the polar equation of the hyperbola: $$r=\frac{m}{1-\sqrt{3} \cos \phi}$$ Convert it to the Cartesian coordinates by using the following relations: $$x=r\cos\phi$$ $$y=r\sin\phi$$ Replacing \(r\) from the polar equation, we get: $$x = \frac{m\cos\phi}{1-\sqrt{3}\cos\phi}$$ $$y = \frac{m\sin\phi}{1-\sqrt{3}\cos\phi}$$ Now, we will eliminate the \(\phi\) parameter. Multiply the second equation by \(\sqrt{3}\) and add both equations: $$x+\sqrt{3}y = m$$ This is the Cartesian equation of the hyperbola.

Find the coordinates of the extremity of the latus rectum

The extremity of the latus rectum is the point where the hyperbola has the maximum value of \(r\). To find this point, find the maximum of the polar equation for \(r\) by taking the derivative with respect to \(\phi\) and equating it to zero, $$\frac{d}{d\phi} \left(\frac{m}{1-\sqrt{3}\cos\phi} \right) = 0$$ This results in, $$\cos\phi=\frac{3}{2\sqrt{3}}$$ Therefore, $$\phi = \arccos\left(\frac{3}{2\sqrt{3}}\right)$$ Now, plug \(\phi\) back into the polar equation: $$x = \frac{m\cos\phi}{1-\sqrt{3}\cos\phi} = \frac{m}{\sqrt{3}}$$ $$y = \frac{m\sin\phi}{1-\sqrt{3}\cos\phi} = \frac{m}{2}$$ So the extremity of the latus rectum is \(\left(\frac{m}{\sqrt{3}},\frac{m}{2}\right)\).

Find the slope of the tangent at the extremity of the latus rectum

Now, take the derivative of the second equation with respect to \(\phi\) and use the chain rule: $$\frac{dy}{d\phi} = \frac{m}{(1-\sqrt{3}\cos\phi)^2}\sin\phi + \frac{m(1-\sqrt{3}\cos\phi)(\sin\phi)}{(1-\sqrt{3}\cos\phi)^2}(-\sqrt{3}\sin\phi)$$ Simplify the equation and evaluate it at the extremity of the latus rectum (plug \(\phi\)): $$\frac{dy}{dx} = \frac{\frac{dy}{d\phi}}{\frac{dx}{d\phi}} = \frac{3}{2\sqrt{3}-3}$$

Determine the angle that the tangent makes with the transverse axis

The angle \(\theta\) that the tangent makes with the transverse axis (x-axis) is given by the following trigonometric relation: $$\tan\theta = \frac{3}{2\sqrt{3}-3}$$ To find the angle, use the arctangent function: $$\theta = \arctan\left(\frac{3}{2\sqrt{3}-3}\right) = 60^{\circ}$$ Thus, the tangent to the hyperbola at the extremity of the latus rectum makes an angle of \(60^{\circ}\) with the transverse axis.

Key concepts

Cartesian Equation of Hyperbola

The Cartesian equation of a hyperbola is a fundamental concept in coordinate geometry. It provides a way to describe the shape of a hyperbola in the familiar x-y coordinate system.

A hyperbola in two dimensions can be represented by the general equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the distances from the center to the vertices and co-vertices, respectively. Converting polar equations like \( r = \frac{m}{1 - \sqrt{3} \cos \phi} \) to Cartesian form involves substituting \( r \cos \phi \) for \(x\) and \( r \sin \phi \) for \(y\), and eliminating the \(\phi\) parameter enables the plotting and analysis in Cartesian coordinates.

This step is crucial as it allows us to exploit powerful algebraic and calculus tools that are better suited for Cartesian forms.

Polar to Cartesian Coordinates Conversion

The translation from polar to Cartesian coordinates is significant in understanding shapes defined in terms of polar equations. Each point in polar coordinates \( (r, \phi) \) corresponds to \( (x, y) \) in Cartesian coordinates by the relations \( x = r \cos \phi \) and \( y = r \sin \phi \).

For students working with geometrical problems, this conversion simplifies the process of analysis, enabling one to deal with more conventional x and y terms. This proves especially useful in calculating slopes, areas, and in this case, confirming the angle a tangent makes with the transverse axis of a hyperbola.

Derivative for Finding Extremity of Latus Rectum

The derivative is a tool that helps us find rates of change and extremities. For hyperbolas, the extremity of the latus rectum is a point of maximum distance from the center along a line parallel to the transverse axis. Calculating this indeed necessitates the use of derivatives.

By differentiating the polar equation of a hyperbola with respect to \(\phi\) and setting it to zero, we can solve for \(\phi\) that gives us the maximum radial distance, \(r\). Solving gives the angle \(\phi\) at which this maximum occurs, and by substituting back into the polar equation, the Cartesian coordinates of the latus rectum can be found.

Tangent Slope Calculation

The slope of a tangent line to a curve at a given point is a measure of how steep the curve is at that point. It's the derivative of the function representing the curve, evaluated at that point. For hyperbolas, calculating the slope involves differentiating the Cartesian equation with respect to \(x\), which often starts with finding \(\frac{dy}{d\phi}\) and \(\frac{dx}{d\phi}\) if the equation is initially in polar form.

The evaluated derivative at a specific point on the curve determines the slope of the tangent. This slope, once found, can be used with trigonometric identities to find the angle that the tangent line makes with the axes, which is a crucial aspect in various geometry and calculus applications.