Problem 196 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(5 x+9=0\) is the directrix of the hyperbola \(16 x^{2}-9 y^{2}=144\), then its corresponding focus is : [April 10, 2019 (II)] (a) \((5,0)\) (b) \(\left(-\frac{5}{3}, 0\right)\) (c) \(\left(\frac{5}{3}, 0\right)\) (d) \((-5,0)\)

Step-by-Step Solution

Answer

The corresponding focus is \(\left(\frac{5}{3}, 0\right)\).

Step 1: Identify the Standard Form of the Hyperbola

The given equation of the hyperbola is \(16x^2 - 9y^2 = 144\). Divide through by 144 to get it in standard form: \[\frac{x^2}{9} - \frac{y^2}{16} = 1\]. This is a horizontal hyperbola centered at the origin (0,0).

Step 2: Determine the Directrix of Hyperbola

The directrix of a hyperbola is given by the equation \(x = \frac{a^2}{e}\) or \(x = -\frac{a^2}{e}\) for a horizontal hyperbola. The given directrix is \(5x + 9 = 0\), solving this gives \(x = -\frac{9}{5}\).

Step 3: Calculate 'a', 'b' and 'e' for the Hyperbola

From the standard form, \(a^2 = 9\) and \(b^2 = 16\). The eccentricity \(e\) is found using \(e = \frac{c}{a}\) where \(c^2 = a^2 + b^2\). Thus, \(c^2 = 9 + 16 = 25\), giving \(c = 5\). Hence, \(e = \frac{5}{3}\).

Step 4: Determine the Corresponding Focus

The foci of the hyperbola are at \((\pm c, 0)\). Here \(c = 5\), hence the possible foci are \((5, 0)\) and \((-5, 0)\). The focus corresponding to the directrix \(x = -\frac{9}{5}\) can be calculated using the formula: \( x = \frac{a^2}{e}\) or \(x = -\frac{a^2}{e}\). Since we calculated \(e = \frac{5}{3}\), thus the x-coordinate of the corresponding focus is \(\pm\frac{a^2 e}{b} = \pm\frac{9 \times \frac{5}{3}}{5}\), simplifying to \(\pm\frac{15}{5} = \pm\frac{5}{3}\). Thus, focus is \(\left(\frac{5}{3}, 0\right)\).

Step 5: Conclusion

Since the calculated x-coordinate is positive, the corresponding focus is \(\left(\frac{5}{3}, 0\right)\).

Key Concepts

Directrix of Hyperbola

In a hyperbola, the directrix is a line that helps in defining the curve's shape and position. It acts as a conceptual reference line from which the deviation of the hyperbola is measured.
For a hyperbola, the distance from any point on the curve to the focus is a fixed multiple of the perpendicular distance from the point to the directrix. This fixed multiple is called the eccentricity, denoted as \( e \).

• For the given hyperbola, \( 16x^2 - 9y^2 = 144 \), the directrix equation \( 5x + 9 = 0 \) simplifies to \( x = -\frac{9}{5} \).
• This directrix is particularly important in determining other properties of the hyperbola, such as its focus and vertices.

Understanding the role of the directrix in relation to the focus helps in grasping the concept of eccentricity as well.

Eccentricity

Eccentricity is a fundamental property of conic sections that defines the curve's "flatness" or "openness." It is denoted by the letter \( e \).
For a hyperbola, the eccentricity \( e \) is always greater than 1. This distinguishes hyperbolas from ellipses, which have eccentricities less than 1, and parabolas with an eccentricity equal to 1.

• To calculate the eccentricity of the given hyperbola \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \), we use the formula \( e = \frac{c}{a} \).
• From the calculations, where \( a^2 = 9 \), \( b^2 = 16 \), and \( c^2 = 25 \), we find \( c = 5 \).
• Thus, the eccentricity is \( e = \frac{5}{3} \), reflecting the hyperbola's openness as it extends horizontally on both sides.

Eccentricity not only helps comprehend the general shape of the hyperbola but also aids in locating its directrix.

Foci of Hyperbola

Foci are critical points in a hyperbola, playing a central role in its geometry. Each hyperbola has two foci, and these are always located along the transverse axis, external to the vertices.
For the hyperbola \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \), the foci provide crucial information about the curve's dimensions. The distance from the center to each focus is denoted by \( c \), which is calculated using \( c^2 = a^2 + b^2 \).

• Here, with \( c = 5 \), the foci are positioned at \( (\pm 5, 0) \).
• The coordinate computation involves matching the focus to the appropriate directrix using the formula \( x = \pm\frac{a^2 e}{b} \).
• This gives us the correct focus in line with the directrix \( x = -\frac{9}{5} \), resulting in coordinates \( \left(\frac{5}{3}, 0\right) \).

The location of the foci assists in understanding the hyperbola's unique reflective properties and its geometric balance.