Problem 151 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A hyperbola having the transverse axis of length \(\sqrt{2}\) has the same foci as that of the ellipse \(3 x^{2}+4 y^{2}=12\), then this hyperbola does not pass through which of the following points? (b) \(\left(-\sqrt{\frac{3}{2}}, 1\right)\) (c) \(\left(1,-\frac{1}{\sqrt{2}}\right)\) (d) \(\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)\)

Step-by-Step Solution

Answer

The hyperbola does not pass through point (d).

Step 1: Determine the Foci of the Ellipse

The equation of the ellipse is \(3x^2 + 4y^2 = 12\). Rearranging it in standard form gives \(\frac{x^2}{4} + \frac{y^2}{3} = 1\). Comparing with \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we have \(a^2 = 4\) and \(b^2 = 3\). The foci of this ellipse are at \((0, \pm c)\) where \(c = \sqrt{a^2 - b^2} = \sqrt{4 - 3} = 1\). So, the foci are at \((0, \pm 1)\).

Step 2: Determine the Equation of the Hyperbola

The hyperbola shares the same foci as the ellipse, so it has foci at \((0, \pm 1)\), meaning \(c = 1\). We know the length of the transverse axis is \(\sqrt{2}\), hence \(2a = \sqrt{2}\) implies \(a = \frac{\sqrt{2}}{2}\). The relationship \(c^2 = a^2 + b^2\) gives \(b^2 = c^2 - a^2 = 1 - \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}\). Therefore, the equation of the hyperbola is \(\frac{x^2}{\left(\frac{\sqrt{2}}{2}\right)^2} - \frac{y^2}{\left(\sqrt{\frac{1}{2}}\right)^2} = 1\), simplifying to \(\frac{x^2}{\frac{1}{2}} - 2y^2 = 1\).

Step 3: Test Each Point in the Hyperbola Equation and Identify Which Does Not Lie on It

1. For the point \((-\sqrt{\frac{3}{2}}, 1)\), plug into the equation: \(\frac{(-\sqrt{\frac{3}{2}})^2}{\frac{1}{2}} - 2 \times 1^2 = \frac{3/2}{1/2} - 2 = 3 - 2 = 1\), it lies on the hyperbola.2. For the point \((1, -\frac{1}{\sqrt{2}})\), plug into the equation: \(\frac{1^2}{\frac{1}{2}} - 2 \times \left(-\frac{1}{\sqrt{2}}\right)^2 = 2 - 1 = 1\), it lies on the hyperbola.3. For the point \((\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}})\), plug into the equation: \(\left(\frac{\sqrt{\frac{3}{2}}}{\sqrt{1/2}}\right)^2 - 2 \times \left(\frac{1}{\sqrt{2}}\right)^2 = 3 - 1 = 2\), it does not satisfy the hyperbola equation.

Key Concepts

Ellipse

An ellipse is a fascinating shape that looks like a stretched circle. It can be defined as the set of all points where the sum of the distances from two fixed points, called "foci," is constant. The general equation of an ellipse is written as \(rac{x^2}{a^2} + rac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

- **Major and Minor Axes:** The longest diameter of the ellipse is called the major axis, while the shortest diameter is the minor axis. These axes are perpendicular and intersect at the center of the ellipse.
- **Role of Foci:** These two fixed points, the foci, are located along the major axis, equidistant from the center. The position of the foci is important for defining the shape and properties of the ellipse.

In the exercise, the ellipse given is \(3x^2 + 4y^2 = 12\). By rearranging it to standard form, \(rac{x^2}{4} + rac{y^2}{3} = 1\), we identify the values \(a^2 = 4\) and \(b^2 = 3\). This helps in finding the foci, which are crucial for further calculations and understanding the related hyperbola.

Hyperbola

A hyperbola looks like two mirrored curves, similar to an open pair of parenthesis. It differs from an ellipse as it describes the set of all points where the difference of the distances to two fixed points, known as "foci," is constant.

- **Standard Form:** The general equation for a hyperbola with a horizontal transverse axis is \(rac{x^2}{a^2} - rac{y^2}{b^2} = 1\). The values \(a\) and \(b\) define the shape and position of the hyperbola's curves. Another format exists for a vertical transverse axis.
- **Foci Relationship:** Like ellipses, hyperbolas have foci, but their determination is based on a slightly different principle, involving the distances being a constant difference rather than sum.

In the exercise, the hyperbola shares the same foci as the ellipse \((0, \pm 1)\). The transverse axis length given is \(\sqrt{2}\), leading to calculations that determine the hyperbola's equation as \(rac{x^2}{rac{1}{2}} - 2y^2 = 1\). Understanding this shape helps in determining which points lie on, and especially do not lie on, it.

Foci

Foci play a vital role in the world of conic sections, such as ellipses and hyperbolas. These are specific points that help define the shape and properties of these curves.

- **Position and Importance:** In an ellipse, the foci are located on the major axis, and they dictate the flattened or elongated nature of the ellipse. The closer the foci are to the center, the more circular the ellipse appears.
- **Contrast in Hyperbolas:** For hyperbolas, the foci stay outside the curves, affecting how far apart they open and influencing the transverse axis. Their contribution lies more in distance rather than positioning within the shape.

In the problem exercise, both the ellipse and the hyperbola share the same foci \(0, \pm 1\). This shared foci makes interesting connections between the ellipse given in the exercise and the related hyperbola, influencing computations like determining what points a curve might pass through.

Transverse Axis

The transverse axis is a term frequently used with hyperbolas. It refers to the line segment that passes through the center of the hyperbola, stretching across from one curve to the other.

- **Understanding the Axis:** In a hyperbola, the transverse axis connects the vertices, the points where each curve is closest to the other. The length of this axis directly affects the shape and breadth of the hyperbola's curves.
- **Calculations and Effects:** The related equation \(2a\) often appears when calculating the length of the transverse axis. It also influences how the other elements of the hyperbola's equation (like \(b\)) are calculated, affecting the overall form.

For the hyperbola in this exercise, the transverse axis length is given as \(\sqrt{2}\). This establishes a basis for determining its equation \(rac{x^2}{rac{1}{2}} - 2y^2 = 1\) and understanding the essential role the transverse axis plays in the hyperbola's geometric description.