Question

If \((\mathrm{x} / 2 \mathrm{a})+[(\mathrm{y} \sqrt{3}) / 2 \mathrm{~b}]=1\) touches the ellipse \(\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)+\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1\), then its eccentric angle \(\theta\) of the contact point is (a) \(0^{\circ}\) (b) \(60^{\circ}\) (c) \(45^{\circ}\) (d) \(90^{\circ}\)

Short answer

The eccentric angle $\theta$ of the contact point is \(60^{\circ}\) (option b).

Step-by-step solution

Rewrite the equation of the line in the standard form

Given the equation of the line as: \((\frac{x}{2a})+(\frac{y\sqrt{3}}{2b})=1\) We will rewrite the equation in the form: \(\frac{x}{a}+\sqrt{3}\frac{y}{b}=2\)

Find the point of contact by substituting the ellipse equation into the line equation

Now, we'll make use of the parametric representation of ellipse: \(x = a\cos{\theta}\), and \(y=b\sin{\theta}\). Substitute \(x = a\cos{\theta}\) and \(y = b\sin{\theta}\) into the line equation: \(\frac{a\cos{\theta}}{a}+\sqrt{3}\frac{b\sin{\theta}}{b}=2\) \(\cos{\theta} + \sqrt{3} \sin{\theta} = 2\)

Determine the value of the eccentric angle θ

Now, we need to find the θ value that will satisfy the equation: \(\cos{\theta} + \sqrt{3} \sin{\theta} = 2\) Apply inverse trigonometric function t from the above equation, \(\theta = \arccos(2 - \sqrt{3}\sin{\theta})\) Now, let's check each option: (a) \(\theta = 0^{\circ}\): \(\cos{0^{\circ}} + \sqrt{3} \sin{0^{\circ}} = 1\), which doesn't satisfy the equation. (b) \(\theta = 60^{\circ}\): \(\cos{60^{\circ}} + \sqrt{3} \sin{60^{\circ}} = 2\), which satisfies the equation. (c) \(\theta = 45^{\circ}\): \(\cos{45^{\circ}} + \sqrt{3} \sin{45^{\circ}} \approx 2.12\), which doesn't satisfy the equation. (d) \(\theta = 90^{\circ}\): \(\cos{90^{\circ}} + \sqrt{3} \sin{90^{\circ}} = \sqrt{3}\), which doesn't satisfy the equation. So, the correct answer is: \( \theta = 60^{\circ}\) (option b).

Key concepts

Understanding the Ellipse

An ellipse is a type of conic section or a curve that you obtain by slicing a cone at an angle with a plane. It's similar in shape to a circle, but stretched out in one direction, making it oval. The equation of an ellipse in its standard form is \[ \left(\frac{x^2}{a^2}\right) + \left(\frac{y^2}{b^2}\right) = 1 \] where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. These axes intersect at the center of the ellipse and are perpendicular to each other.

In an ellipse, similar to a circle, any straight line drawn from one point on the curve to another will pass within the shape, showcasing the 'closed curve' property. One of the fascinating things about an ellipse is that it has two focal points, and the sum of the distances from any point on the ellipse to these focal points is constant. This constant is larger than the distance between the focal points, which defines the stretched nature of an ellipse compared to a circle.

When we're dealing with JEE Maths problems involving ellipses, it's crucial to understand the significance of these elements, as they influence the curve’s geometry and are often key to solving related problems.

Parametric Representation of Ellipse

The parametric form of an ellipse is an alternative way of describing the ellipse using a single parameter, usually denoted as \(\theta\), which is known as the eccentric angle. In parametric terms, the ellipse's equation can be expressed as: \[ x = a\cos(\theta) \] \[ y = b\sin(\theta) \]

This representation ties every point on the ellipse to a specific angle, thereby simplifying many calculus and algebra problems. For example, if you wanted to find the point on the ellipse that the line \((x/2a) + (y\sqrt{3}/2b) = 1\) touches, you'd substitute the parametric equations into the line’s equation. The result will give you the exact point of contact in terms of the angle \(\theta\), making this approach incredibly powerful for solving JEE Maths problems involving ellipses.

In practice, the parametric equations help us visualize how the ellipse is traced as \(\theta\) varies from 0 to \(2\pi\). Every value of \(\theta\) corresponds to a unique point on the ellipse, which can be particularly useful for graphing or integrating along the curve.

Eccentric Angle in Ellipse

The eccentric angle \(\theta\) in the context of an ellipse is the angle formed by the line segment from the center of the ellipse to a point on the ellipse and the major axis. It's a parameter in the parametric equations of the ellipse. Understanding the eccentric angle is valuable, especially when finding the points of tangency for lines that touch the ellipse, which is a common type of problem in JEE Maths.

For instance, to solve the problem of finding the eccentric angle \(\theta\) of a contact point where a line touches the ellipse, we would use the parametric equations of the ellipse, substitute them into the equation of the line, and solve for \(\theta\). As shown in our example, when solving the equation \(\cos(\theta) + \sqrt{3} \sin(\theta) = 2\), we find that \(\theta = 60^\circ\) fulfills this condition.

Remembering that the eccentric angles correspond to specific points on the ellipse is crucial for not only finding points of contact but also for many other applications involving arc length, area, or the ellipse's properties in physics and engineering problems. By mastering the use of the eccentric angle, students can significantly enhance their problem-solving toolkit for tackling advanced mathematics.