Problem 189 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(e_{1}\) and \(e_{2}\) be the eccentricities of the ellipse, \(\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b<5)\) and the hyperbola, \(\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1\) respectively satisfying \(e_{1} \mathrm{e}_{2}=1\). If \(\alpha\) and \(\beta\) are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair \((\alpha, \beta)\) is equal to : (a) \((8,12)\) (b) \(\left(\frac{20}{3}, 12\right)\) (c) \(\left(\frac{24}{5}, 10\right)\) (d) \((8,10)\)

Step-by-Step Solution

Answer

The ordered pair \((\alpha, \beta)\) is (d) \((8, 10)\).

Step 1: Identify Eccentricities

For the ellipse given by the equation \( \frac{x^{2}}{25} + \frac{y^{2}}{b^{2}} = 1 \), the eccentricity \( e_1 \) is defined as \( e_1 = \sqrt{1 - \frac{b^{2}}{25}} \). For the hyperbola given by the equation \( \frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = 1 \), the eccentricity \( e_2 \) is defined as \( e_2 = \sqrt{1 + \frac{b^{2}}{16}} \).

Step 2: Use the Eccentricity Product Condition

We are given that the product of the eccentricities satisfies the equation \( e_1 e_2 = 1 \). Plug in the formulas for \( e_1 \) and \( e_2 \) to get \( \sqrt{1 - \frac{b^{2}}{25}} \cdot \sqrt{1 + \frac{b^{2}}{16}} = 1 \).

Step 3: Solve the Eccentricity Equation

Squaring both sides of \( \sqrt{1 - \frac{b^{2}}{25}} \cdot \sqrt{1 + \frac{b^{2}}{16}} = 1 \), we have \( \left(1 - \frac{b^{2}}{25}\right) \left(1 + \frac{b^{2}}{16}\right) = 1 \). Expand and simplify: \( 1 + \frac{b^{2}}{16} - \frac{b^{2}}{25} - \frac{b^{4}}{400} = 1 \). This simplifies to \( \frac{9b^2}{400} - \frac{b^4}{400} = 0 \), which simplifies further to \( b^2(9 - b^2) = 0 \). Hence, \( b^2 = 9 \) (since \( b < 5 \)).

Step 4: Calculate the Foci Distances

With \( b^2 = 9 \), calculate the distance between the foci of the ellipse: \( \alpha = 2ae_1 = 2 \times 5 \times \frac{4}{5} = 8 \). Similarly, for the hyperbola, the distance between the foci is \( \beta = 2ae_2 = 2 \times 4 \times \frac{5}{4} = 10 \).

Step 5: Determine the Correct Option

The calculated distances for the foci of the ellipse and hyperbola are \( \alpha = 8 \) and \( \beta = 10 \). The ordered pair \((\alpha, \beta)\) is therefore \((8, 10)\). Comparing with the options, the correct answer is (d) \((8, 10)\).

Key Concepts

Ellipse

An ellipse is a fascinating shape in geometry that looks like an elongated circle. The standard form of an ellipse's equation that we often work with is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. If \(a \geq b\), the ellipse is oriented horizontally, and if \(b \geq a\), it is oriented vertically.
The space where the ellipse stretches the most is along its major axis, and the shortest span is across its minor axis. The sum of the distances from any point on the ellipse to the two foci, another critical concept, is always constant. This unique feature differentiates ellipses from other conic sections like hyperbolas and parabolas.

• An ellipse can be thought of as a stretched circle, where each point adheres to the rule of equal summed distances to two fixed points called foci.
• The more stretched or elongated the ellipse, the higher its eccentricity.

Hyperbola

A hyperbola is another type of conic section, distinct from the circle and ellipse due to its open curve formation. Its equation can be written as \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), making it immediately recognizable by the minus sign.
Unlike an ellipse, which loops back around onto itself, a hyperbola consists of two distinct branches. These branches mirror each other and open outward, either horizontally or vertically, depending on the equation's orientation.

• The hyperbola opens horizontally if the \(x\) term comes first and vertically if the \(y\) term comes first in the equation.
• Each branch of the hyperbola gets closer and closer to a pair of lines called asymptotes, but never actually touches them.
• As with an ellipse, a hyperbola has two foci, but the sum of the distances to the foci from any point on the hyperbola is not constant. Instead, for any point on the hyperbola, the absolute difference in distances to the foci is constant.

Eccentricity

Eccentricity is a term that helps to differentiate conic sections from one another. It signifies how much a conic section deviates from being circular. For an ellipse, the eccentricity \(e\) is calculated as \(e = \sqrt{1 - \frac{b^2}{a^2}}\) and always has a value between 0 and 1. The closer the value is to 0, the more circular the ellipse looks.
For hyperbolas, the eccentricity is given by \(e = \sqrt{1 + \frac{b^2}{a^2}}\), and it will be greater than 1, indicating its open curve.

• A circle is a special kind of ellipse with an eccentricity of 0.
• Higher eccentricity values in ellipses indicate a more elongated shape.
• Hyperbolas with large eccentricities become more stretched and narrow.

Eccentricity plays a crucial role in the clarity of shapes, defining whether they are ellipses or hyperbolas, and it provides essential insight into the elements like foci spacing and the type of conic section.

Foci

The concept of foci involves two points that are integral to defining conic sections. The foci of an ellipse and a hyperbola have distinct roles in determining the shape and properties of these curves.
In an ellipse, the foci are two fixed points inside the shape. For any point on the ellipse, the total distance to both foci is always constant. This is a primary factor leading to the unique ellipse shape. If you think of the ellipse like an elastic band stretched around two pins (the foci), this band will form the ellipse shape allowing for this total fixed distance.

• In an ellipse, as foci move closer, the shape becomes more circular.
• When the foci coincide at the center, the ellipse becomes a circle.

In hyperbolas, the foci also play a pivotal role but with a different characteristic. For points on a hyperbola, the absolute difference in distances to the foci remains constant rather than their sum.

• In hyperbolas, the foci are outside the curve.
• The further apart the foci are, the more open the hyperbola branches become.

Overall, when trailing foci whether for an ellipse or hyperbola, these points guide the formation and properties of the conic section, fitting essential mathematical and geometric principles.