Problem 93 - A Complete Resource Book in Mathematics for JEE Main

Question

A focus of an ellipse is at the origin. The directrix is the line \(x=4\) and the eccentricity is \(\frac{1}{2}\), Then the length of the semi-major axis is (A) \(\frac{8}{3}\) (B) \(\frac{2}{3}\) (C) \(\frac{4}{3}\) (D) \(\frac{5}{3}\)

Step-by-Step Solution

Answer

The length of the semi-major axis is \(\frac{8}{3}\). (Option A)

Step 1: Understand the problem requirements

We need to find the length of the semi-major axis of an ellipse given a focus at the origin, a directrix line at \(x=4\), and an eccentricity \(e=\frac{1}{2}\).

Step 2: Recall the ellipse properties

An ellipse with focus at the origin has the general property that for any point \((x,y)\) on the ellipse, the distance to the focus \((0,0)\) is equal to \(e\) times the distance to the directrix line \(x=4\).

Step 3: Apply the ellipse condition

For a point \((x,y)\) on the ellipse: the distance to the focus is \(\sqrt{x^2+y^2}\) and the distance to the directrix \(x=4\) is \(|x-4|\). The relationship is \(\sqrt{x^2 + y^2} = \frac{1}{2} |x-4|\).

Step 4: Determine 'a' using the center-focus relationship

Using the standard equation of an ellipse centered at the origin with a horizontal semi-major axis, the eccentricity \(e\) is given by \(e=\frac{c}{a}\), where \(c\) is the distance from the center to the focus. Given \(c=0\) since focus is at origin, semi-major axis length is \(a=\frac{distance\ to\ directrix \cdot e}{e^2-1}\).

Step 5: Calculate the semi-major axis

Given the directrix distance is 4, \(a = \frac{4 \times \frac{1}{2}}{1-(\frac{1}{2})^2} = \frac{8}{3}\). Thus, the length of the semi-major axis is \(\frac{8}{3}\).

Key Concepts

Focus and directrix of ellipse

An ellipse is a fascinating shape with unique properties. It has two special points called foci (plural of focus), and each point on the ellipse has a particular age-old relationship with these foci. For an ellipse in our exercise, the focus is located at the origin, which is point (0,0). The directrix, on the other hand, is a fixed line, and for our ellipse, it is represented by the equation \(x=4\). This means it's a vertical line positioned at four units to the right of the origin.The primary property of an ellipse should be noted here: for any point on the ellipse, the ratio of its distance to the focus and its distance to the directrix is constant. This constant ratio is known as the eccentricity, which we'll delve into shortly.

Eccentricity of ellipse

Eccentricity, often represented by the symbol \(e\), is a measure of how elongated an ellipse is. For a perfectly round circle, the eccentricity is 0, while for our ellipse in the exercise, it is \(\frac{1}{2}\). This tells us that our ellipse is not a perfect circle, but it's also not overly elongated.This value of eccentricity, \(\frac{1}{2}\), implies a specific relationship in the ellipse. If you pick any point on the ellipse, the distance to the focus at (0,0) is always half the distance to the directrix \(x=4\). This helps us understand the layout and dimensions of the ellipse beyond just its visual appearance. Eccentricity is essential in determining how stretched out the ellipse is and influences the calculation of other ellipse properties.

Semi-major axis calculation

The semi-major axis of an ellipse is one of its longest radii, reaching from its center to the edge. Calculating it requires understanding the relationships between the ellipse's components: focus, directrix, and eccentricity. In this exercise, we know:

• The focus of the ellipse.
• The directrix line \(x=4\).
• The eccentricity \(e=\frac{1}{2}\).

For our ellipse centered at the origin, the relationship between the center and focus is captured in the formula \[ e = \frac{c}{a} \]where \(c\) is the distance from center to focus. Given that our focus is the origin, \(c=0\), our task is to find \(a\), the semi-major axis. Since the distance to the directrix is 4, we apply the formula\[ a = \frac{\text{distance to directrix} \times e}{1-e^2} \]Plugging in the given values, we find\[ a = \frac{4 \times \frac{1}{2}}{1-(\frac{1}{2})^2} = \frac{8}{3} \]Thus, the semi-major axis length of the ellipse is \(\frac{8}{3}\). This calculation is crucial to fully understanding the size and shape of the ellipse.