NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.4

Understanding the Hyperbola and its Standard Equations

A hyperbola is defined as the set of all points in a plane, the difference of whose distances from two fixed points (called foci) is a constant. This is in contrast to an ellipse, where the sum of the distances is constant. Unlike an ellipse where the orientation is determined by which denominator is larger, for a hyperbola, the orientation is determined strictly by which term is positive.

There are two standard orientations for a hyperbola centered at the origin $(0,0)$:

1. Horizontal Transverse Axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Here, the transverse axis lies along the x-axis, the vertices are $(\pm a, 0)$, and the foci are $(\pm c, 0)$.
2. Vertical Transverse Axis: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. Here, the transverse axis lies along the y-axis, the vertices are $(0, \pm a)$, and the foci are $(0, \pm c)$.

For both types, the semi-transverse axis has length $a$, the semi-conjugate axis has length $b$, and the distance of the foci from the center is $c$, which obeys the fundamental relationship $c^2 = a^2 + b^2$. The eccentricity is given by $e = \frac{c}{a}$, and because $c > a$, the eccentricity of a hyperbola is always greater than 1 ($e > 1$). The length of the latus rectum is calculated using the formula $\frac{2b^2}{a}$.

Key Terminology of a Hyperbola

Transverse Axis

The line segment passing through the two foci and vertices of the hyperbola. Its total length is 2a.

Conjugate Axis

The line segment perpendicular to the transverse axis, passing through the center. Its total length is 2b.

Eccentricity (e)

The ratio of the distance of a point on the hyperbola from the focus to its distance from the directrix, mathematically represented as e = c/a.

Latus Rectum

A chord passing through either of the foci and perpendicular to the transverse axis. Its length is 2b²/a.

Step-by-Step Process to Analyze any Hyperbola Equation

1. Step 1: Convert to Standard Form — Ensure the RHS is equal to 1. If it is not, divide both sides of the equation by the constant on the RHS.
2. Step 2: Identify the Orientation — Look at which squared term is positive. If x² is positive, it is a horizontal hyperbola. If y² is positive, it is a vertical hyperbola.
3. Step 3: Extract 'a' and 'b' — The denominator of the positive term is a², and the denominator of the negative term is b².
4. Step 4: Compute 'c' — Use the formula c² = a² + b² to calculate the focal distance, c.
5. Step 5: Write Down Key Coordinates and Measures — Determine the vertices, foci, eccentricity (e = c/a), and length of the latus rectum (2b²/a) based on the orientation.

CBSE Board Exam Traps & Tips

Trap 1: The 'a' and 'b' Switch Confusion. In ellipses, $a^2$ is always the larger denominator. However, in hyperbolas, $a^2$ is always the denominator under the positive term, regardless of whether it is larger or smaller than $b^2$. Do not assume the larger number is always $a^2$!

Trap 2: Formula Sign Error. Remember that for an ellipse, $b^2 = a^2(1 - e^2)$ which means $c^2 = a^2 - b^2$. But for a hyperbola, $c^2 = a^2 + b^2$. Keep these formulas separate to avoid losing easy marks on your exams.

Practice Questions with Solutions

• Q: Find the coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$. A: Step 1: Compare the given equation with the standard equation of a horizontal hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Here, $a^2 = 9$ and $b^2 = 16$. This gives $a = 3$ and $b = 4$. Step 2: Calculate $c$ using the relation $c^2 = a^2 + b^2$. $c^2 = 9 + 16 = 25 \implies c = 5$. Step 3: State the parameters: Vertices: $(\pm a, 0) = (\pm 3, 0)$ Foci: $(\pm c, 0) = (\pm 5, 0)$ Eccentricity: $e = \frac{c}{a} = \frac{5}{3}$ Length of latus rectum: $\frac{2b^2}{a} = \frac{2(16)}{3} = \frac{32}{3}$. Final answer: Vertices are $(\pm 3, 0)$, Foci are $(\pm 5, 0)$, Eccentricity is $\frac{5}{3}$, and Length of Latus Rectum is $\frac{32}{3}$.
• Q: Find the coordinates of the foci, vertices, eccentricity, and length of the latus rectum for $9y^2 - 4x^2 = 36$. A: Step 1: Divide both sides of the equation by 36 to make the RHS equal to 1: $\frac{9y^2}{36} - \frac{4x^2}{36} = 1 \implies \frac{y^2}{4} - \frac{x^2}{9} = 1$. Step 2: This is a vertical hyperbola of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. Here, $a^2 = 4 \implies a = 2$, and $b^2 = 9 \implies b = 3$. Step 3: Find $c$: $c^2 = a^2 + b^2 = 4 + 9 = 13 \implies c = \sqrt{13}$. Step 4: Determine the properties: Vertices: $(0, \pm a) = (0, \pm 2)$ Foci: $(0, \pm c) = (0, \pm \sqrt{13})$ Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{13}}{2}$ Length of latus rectum: $\frac{2b^2}{a} = \frac{2(9)}{2} = 9$. Final answer: Vertices: $(0, \pm 2)$, Foci: $(0, \pm \sqrt{13})$, Eccentricity: $\frac{\sqrt{13}}{2}$, Length of Latus Rectum: $9$.
• Q: Find the equation of the hyperbola satisfying the given conditions: Vertices $(\pm 2, 0)$, Foci $(\pm 3, 0)$. A: Step 1: Since the vertices and foci lie on the x-axis, the hyperbola is horizontal. Its standard equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Step 2: From the coordinates, the vertices are $(\pm a, 0) = (\pm 2, 0)$, so $a = 2 \implies a^2 = 4$. The foci are $(\pm c, 0) = (\pm 3, 0)$, so $c = 3 \implies c^2 = 9$. Step 3: Use the relationship $c^2 = a^2 + b^2$ to find $b^2$: $9 = 4 + b^2 \implies b^2 = 5$. Step 4: Substitute $a^2$ and $b^2$ into the standard equation: $\frac{x^2}{4} - \frac{y^2}{5} = 1$. Final answer: The equation is $\frac{x^2}{4} - \frac{y^2}{5} = 1$.
• Q: Find the equation of the hyperbola satisfying the conditions: Foci $(0, \pm 13)$, conjugate axis of length $24$. A: Step 1: The foci lie on the y-axis, indicating a vertical hyperbola. Its equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. Step 2: Foci are $(0, \pm c) = (0, \pm 13)$, so $c = 13 \implies c^2 = 169$. Step 3: The length of the conjugate axis is $2b = 24$, which gives $b = 12 \implies b^2 = 144$. Step 4: Use the relation $c^2 = a^2 + b^2$ to solve for $a^2$: $169 = a^2 + 144 \implies a^2 = 25$. Step 5: Write the final equation: $\frac{y^2}{25} - \frac{x^2}{144} = 1$. Final answer: The equation is $\frac{y^2}{25} - \frac{x^2}{144} = 1$.

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