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Conic Sections - Hyperbola

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition: A hyperbola is defined as the set of all points in a plane where the absolute difference of the distances from two fixed points (called foci) is a constant. Visually, a hyperbola consists of two separate, non-connecting curves known as branches, which are mirror images of each other across the axes.

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Transverse and Conjugate Axes: The line segment passing through the two foci is called the transverse axis. Its length is 2a2a. The line segment perpendicular to the transverse axis and passing through the center is the conjugate axis, with length 2b2b. Visually, the transverse axis connects the two vertices of the branches, while the conjugate axis sits in the 'gap' between the two branches.

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Standard Form (Horizontal): The equation x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 represents a hyperbola with its transverse axis along the xx-axis. Visually, the branches of this hyperbola open to the left and right. The vertices are located at (±a,0)(\pm a, 0) and the foci are at (±c,0)(\pm c, 0).

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Standard Form (Vertical): The equation y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 represents a hyperbola with its transverse axis along the yy-axis. Visually, the branches open upwards and downwards. The vertices are at (0,±a)(0, \pm a) and the foci are at (0,±c)(0, \pm c). Note that in a hyperbola, the positive term determines the orientation, not the size of the denominator.

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Eccentricity (ee): This is the ratio of the distance from the center to a focus (cc) to the distance from the center to a vertex (aa), given by e=cae = \frac{c}{a}. For every hyperbola, e>1e > 1. Visually, as ee increases, the branches of the hyperbola appear 'wider' or flatter.

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Fundamental Relationship: The constants aa (semi-transverse axis), bb (semi-conjugate axis), and cc (distance from center to focus) are related by the equation c2=a2+b2c^2 = a^2 + b^2. This implies that cc is always greater than aa, meaning the foci are always located further from the center than the vertices, effectively 'inside' the curve of the branches.

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Latus Rectum: This is the line segment perpendicular to the transverse axis, passing through one of the foci, with endpoints on the hyperbola branches. The length of the latus rectum is given by 2b2a\frac{2b^2}{a}. Visually, it measures the vertical or horizontal width of the hyperbola at the focus point.

📐Formulae

Standard Equation (Horizontal): x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

Standard Equation (Vertical): y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

Relationship between constants: c2=a2+b2c^2 = a^2 + b^2 (where c=a2+b2c = \sqrt{a^2 + b^2})

Eccentricity: e=ca=a2+b2ae = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a}

Length of Transverse Axis: 2a2a

Length of Conjugate Axis: 2b2b

Length of Latus Rectum: 2b2a\frac{2b^2}{a}

Coordinates of Foci: (±c,0)(\pm c, 0) for horizontal; (0,±c)(0, \pm c) for vertical

Coordinates of Vertices: (±a,0)(\pm a, 0) for horizontal; (0,±a)(0, \pm a) for vertical

💡Examples

Problem 1:

Find the coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the hyperbola given by the equation x216−y29=1\frac{x^2}{16} - \frac{y^2}{9} = 1.

Solution:

  1. Comparing the given equation x216−y29=1\frac{x^2}{16} - \frac{y^2}{9} = 1 with the standard form x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, we get a2=16a^2 = 16 and b2=9b^2 = 9. Thus, a=4a = 4 and b=3b = 3.
  2. Calculate cc using c2=a2+b2c^2 = a^2 + b^2: c2=16+9=25c^2 = 16 + 9 = 25, so c=5c = 5.
  3. Vertices: Since the xx-term is positive, the transverse axis is on the xx-axis. Vertices are (±a,0)=(±4,0)(\pm a, 0) = (\pm 4, 0).
  4. Foci: The foci are (±c,0)=(±5,0)(\pm c, 0) = (\pm 5, 0).
  5. Eccentricity: e=ca=54=1.25e = \frac{c}{a} = \frac{5}{4} = 1.25.
  6. Length of Latus Rectum: 2b2a=2(9)4=184=4.5\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{18}{4} = 4.5.

Explanation:

This problem requires identifying the orientation of the hyperbola from its equation, determining the constants aa and bb, and then using standard formulas to find the geometric properties.

Problem 2:

Find the equation of the hyperbola whose foci are at (0,±13)(0, \pm 13) and the length of the conjugate axis is 24.

Solution:

  1. The foci are on the yy-axis at (0,±13)(0, \pm 13), so the hyperbola is vertical and c=13c = 13.
  2. The length of the conjugate axis is 2b=242b = 24, which gives b=12b = 12 and b2=144b^2 = 144.
  3. Use the relationship c2=a2+b2c^2 = a^2 + b^2 to find a2a^2: 132=a2+122⇒169=a2+144⇒a2=169−144=2513^2 = a^2 + 12^2 \Rightarrow 169 = a^2 + 144 \Rightarrow a^2 = 169 - 144 = 25.
  4. The standard equation for a vertical hyperbola is y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1.
  5. Substituting a2a^2 and b2b^2: y225−x2144=1\frac{y^2}{25} - \frac{x^2}{144} = 1.

Explanation:

To find the equation, we first determine the orientation from the foci. Then we find a2a^2 and b2b^2 using the given conjugate axis length and the focus distance formula.