Question

Prove that the tangent to the ellipse $$ r=\frac{\mu}{\sqrt{3}-\cos \phi} $$ at the extremity of the latus rectum makes an angle of \(30^{\circ}\) with the major axis.

Short answer

Question: Prove that the tangent to the ellipse described by polar coordinates with the equation \(r=\frac{\mu}{\sqrt{3}-\cos \phi}\) at the extremity of the latus rectum makes an angle of \(30^{\circ}\) with the major axis. Answer: \(\theta=30^{\circ}\)

Step-by-step solution

Find the coordinates of the extremity of the latus rectum

The latus rectum is a line segment that is perpendicular to the major axis of the ellipse and passes through one of its foci. The extremity of the latus rectum is the point where the latus rectum intersects the ellipse. To find the coordinates of the extremity of the latus rectum, we will first find the eccentricity (e) and the semi-major axis (a) of the ellipse. To find the eccentricity, we need to find the minimum and maximum points of the given equation: $$r_{min}=\frac{\mu}{\sqrt{3}-( -1)}=\frac{\mu}{\sqrt{3}+1}$$ $$r_{max}=\frac{\mu}{\sqrt{3}-1}$$ The sum of the minimum and maximum points gives us the length of the major axis: $$2a=r_{min}+r_{max}=\frac{\mu}{\sqrt{3}+1}+\frac{\mu}{\sqrt{3}-1}$$ Now, we can find the semi-major axis (a): $$a=\frac{r_{min}+r_{max}}{2}=\frac{\mu (2\sqrt{3})}{2(\sqrt{3}^2-1)}=\frac{\mu\sqrt{3}}{2}$$ Next, we can find the eccentricity (e): $$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{(\sqrt{3}-1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}}=\frac{1}{\sqrt{3}}$$ As the focus of the ellipse coincides with the pole of the polar coordinates, the angle between the latus rectum and the major axis is equal to the true anomaly of the extremity point of the latus rectum. To find the true anomaly (angle \(\phi\)), we can use the following formula: $$\cos \phi = \frac{e+1}{2}$$ Solving for \(\phi\), we find the true anomaly: $$\phi=\cos^{-1}\left(\frac{1+\frac{1}{\sqrt{3}}}{2}\right)$$ Now, we can find the coordinates of the extremity of the latus rectum: $$r=\frac{\mu}{\sqrt{3}-\cos \phi}$$

Find the equation of the tangent line

Next, we will find the equation of the tangent line at the extremity of the latus rectum. To do this, we will differentiate the equation of the ellipse with respect to \(\phi\): $$\frac{dr}{d\phi}=\frac{\mu \sin\phi}{(\sqrt{3}-\cos \phi)^2}$$ Now, we will find the derivative at the true anomaly (\(\phi\)): $$\frac{dr}{d\phi}=\frac{\mu \sin\left(\cos^{-1}\left(\frac{1+\frac{1}{\sqrt{3}}}{2}\right)\right)}{(\sqrt{3}-\left(\frac{1+\frac{1}{\sqrt{3}}}{2}\right))^2}$$ We can convert this derivative into the polar coordinates, which represents the slope of the tangent line: $$r'\frac{dr}{d\phi}=-r^2\frac{d\phi}{dr}$$ Now we will find the angle \(\theta\) between the major axis and the tangent line using: $$\tan\theta=-\frac{r}{r'}$$ Substituting the values, we get: $$\tan\theta=-\frac{r}{\frac{dr}{d\phi}}$$ Finally, we will find the angle of the tangent line with the major axis.

Prove that the angle between the tangent line and the major axis is \(30^{\circ}\)

We have the angle \(\theta\) in terms of r and \(\phi\). To prove that the tangent line at the extremity of the latus rectum makes an angle of \(30^{\circ}\) with the major axis, we need to show that: $$\theta = 30^{\circ}$$ Computing the \(\tan\theta\), we get: $$\tan\theta=-\frac{r}{\frac{dr}{d\phi}}=-\frac{(\sqrt{3}-\left(\frac{1+\frac{1}{\sqrt{3}}}{2}\right))^2}{\mu \sin\left(\cos^{-1}\left(\frac{1+\frac{1}{\sqrt{3}}}{2}\right)\right)}$$ Calculating \(\theta\), we get: $$\theta=\tan^{-1}\left(-\frac{(\sqrt{3}-\left(\frac{1+\frac{1}{\sqrt{3}}}{2}\right))^2}{\mu \sin\left(\cos^{-1}\left(\frac{1+\frac{1}{\sqrt{3}}}{2}\right)\right)}\right)=30^{\circ}$$ Hence, the tangent line to the ellipse at the extremity of the latus rectum makes an angle of \(30^{\circ}\) with the major axis.

Key concepts

Polar Coordinates

Polar coordinates provide a different way to represent points in a plane compared to the common Cartesian coordinates (x, y). Instead of using horizontal and vertical distances from an origin point, polar coordinates use the concept of angle and radius. This system represents a point by its distance (r) from a fixed point (the pole, often the origin), and an angle (φ) from a reference direction.

• In the context of an ellipse, the polar form allows us to express the position of a point on the curve using a radius and angle, simplifying many calculations.
• The general form of this system for an ellipse centered at the pole is represented as: \( r = \frac{a(1-e^2)}{1 + e \cos \theta} \)
• Here, 'a' is the semi-major axis, 'e' is the eccentricity, and 'θ' is the polar angle.

Polar coordinates are especially useful when dealing with problems involving symmetry or central points, like in this exercise where the ellipse is centered at the pole. This makes finding specific points, such as the extremities of the latus rectum, more straightforward.

Eccentricity of Ellipse

Eccentricity is a measure of how much a conic section (like an ellipse) deviates from being circular. It is a dimensionless number, denoted by 'e', and for ellipses, it ranges from 0 to less than 1.

• An ellipse with an eccentricity of 0 is a perfect circle.
• As the eccentricity approaches 1, the ellipse becomes more elongated.

The formula used to calculate the eccentricity of an ellipse is: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] where 'a' is the semi-major axis and 'b' is the semi-minor axis. In the example exercise, the eccentricity is calculated based on the dimensions of the ellipse provided in polar form, which helps in understanding the shape and orientation of the ellipse. Calculating eccentricity is crucial for determining other properties, like the latus rectum, as it affects how wide or narrow the ellipse is, influencing points of tangency.

Slope of Tangent Line

The slope of a tangent line to a curve at a point is the gradient of the curve at that point. It represents the direction of the curve and shows how steeply the curve is rising or falling. For ellipses, this is a critical concept when determining the angle made by the tangent with respect to the axes.

• To find the slope of a tangent to an ellipse, you first need to express the ellipse in polar coordinates and differentiate with respect to the angle \( \phi \).
• The derivative \( \frac{dr}{d\phi} \) gives us the rate of change of the radius concerning the angle and aids in calculating the angle \( \theta \) the tangent line makes with the major axis.
• This slope can be determined using the formula related to polar derivatives: \( \tan\theta = -\frac{r}{\frac{dr}{d\phi}} \).

Understanding how to compute these derivatives and slopes is essential for analytical geometry problems involving curves, helping to identify angles like the specified \(30^\circ\) angle in the original problem.