Question

A hyperbola, having the transverse axis of length \(2 \sin \theta\) is confocal with the ellipse \(3 \mathrm{x}^{2}+4 \mathrm{y}^{2}=12\). Then its equation is (a) \(x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1\) (b) \(x^{2} \sec ^{2} \theta-y^{2} \operatorname{cosec}^{2} \theta=1\) (c) \(x^{2} \sin ^{2} \theta-y^{2} \cos ^{2} \theta=1\) (d) \(x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta=1\)

Short answer

The short answer is: \(x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta=1\).

Step-by-step solution

Express the ellipse in standard form

Divide ellipse equation by 12 to obtain standard form: \[ \frac{x^2}{(12/3)} + \frac{y^2}{(12/4)} = 1\] which simplifies to: \[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]

Calculate ellipse foci

Given the ellipse equation: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] We can find the distance between the foci for ellipse: \[c = \sqrt{a^2 - b^2}\] Substitute \(a^2=4\) and \(b^2=3\) and calculate the distance between foci: \[c= \sqrt{4-3}=\sqrt{1}=1\] Since the ellipse is horizontal, ellipse foci are (±1,0).

Calculate transverse axis and semi-transverse axis for hyperbola

We are given the length of the transverse axis of the hyperbola as \(2\sin\theta\). The semi-transverse axis (a) for the hyperbola is then: \[a = \sin \theta\]

Calculate semi-conjugate axis for hyperbola and equation

Since hyperbola and ellipse are confocal, they share the same foci. The distance between foci for hyperbola is also c = 1. Then, we can use this relation for hyperbola: \[c^2 = a^2 + b^2\] where a is the semi-transverse axis, b is the semi-conjugate axis, and c is the distance between the foci. Substitute given values: \[1^2 = (\sin \theta)^2 + b^2\] Solve for \(b^2\): \[b^2 = 1 - (\sin \theta)^2 = \cos^2\theta \]

Write equation of hyperbola

The standard form equation of a horizontal hyperbola is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] Substitute the values of \(a^2\) and \(b^2\) that we found earlier: \[ \frac{x^2}{(\sin^2\theta)} - \frac{y^2}{(\cos^2\theta)} = 1 \] This matches option (d): \(x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta=1\)

Key concepts

Transverse Axis

The transverse axis is a crucial element in understanding both ellipses and hyperbolas. It is the line that runs through the center and the foci of an ellipse or a hyperbola, and its length is one of the defining features of these curves.

For an ellipse, the transverse axis is the longest diameter and lies along the major axis. In contrast, for a hyperbola, the transverse axis is found along its opening, between the vertices. The length of the transverse axis provides critical information needed to calculate the equations that represent these shapes.

In terms of the shared foci condition known as 'confocal,' it means that the transverse axis of the hyperbola and the major axis of the ellipse are aligned with each other, possessing the same central point and foci. This shared foci scenario plays an important part in finding the relationship between the sizes of a confocal ellipse and hyperbola.

Ellipse Standard Form

The standard form of an ellipse's equation is an essential building block for solving geometric problems related to ellipses. It is represented by:

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
Where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively. For a horizontal ellipse, where the transverse axis lies along the x-axis, \(a\) would be greater than \(b\) and for a vertical ellipse, it's the opposite.

If we're provided an ellipse equation like \(3x^2 + 4y^2 = 12\), we would first normalize it by dividing each term by 12 to achieve the standard form. This makes it easier to identify and compare the properties of the ellipse, such as its axes lengths, center, and foci, with those of other conic sections such as hyperbolas, particularly when they are confocal.

Hyperbola Equation

The equation of a hyperbola is another key concept and is similar in form to that of an ellipse, but with a vital difference that reflects its distinct shape. A hyperbola's standard equation takes one of two forms depending on its orientation:

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\] (horizontal hyperbola)
or
\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\] (vertical hyperbola)

Here, \(a\) is the length of the semi-transverse axis, while \(b\) is the length of the semi-conjugate axis. It's important to note that the transverse axis corresponds to the axis that 'opens' the hyperbola, whereas the conjugate axis is perpendicular to the transverse axis.

When a hyperbola is confocal with an ellipse, they share the same foci, which means that the sum and differences of the squares of the lengths of their axes are related. This relationship helps us write the hyperbola's equation when given certain parameters, like the transverse axis length and the shared foci, which will guide us to the correct form of the hyperbola's equation in the context of the problem.