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Conic Sections - Ellipse

Grade 11CBSE

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

🔑Concepts

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An ellipse is defined as the set of all points in a plane such that the sum of their distances from two fixed points, called foci (F1F_1 and F2F_2), is a constant value. Visually, this creates a smooth, closed curve that looks like a stretched circle, where the sum of the 'focal radii' for any point on the boundary remains equal to the length of the major axis (2a2a).

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The Major Axis is the longest diameter of the ellipse, passing through both foci and the center. Its endpoints are the vertices. The Minor Axis is the shortest diameter, passing through the center perpendicular to the major axis. In a horizontal ellipse, the major axis lies along the xx-axis, making the shape wider than it is tall; in a vertical ellipse, it lies along the yy-axis, making it taller than it is wide.

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The Center of the ellipse is the midpoint of the line segment joining the two foci. It is the point of symmetry for the entire figure. The distance from the center to either vertex on the major axis is denoted by aa, the distance to the endpoints of the minor axis is bb, and the distance to each focus is cc.

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The Relationship between semi-axes is given by the equation c2=a2−b2c^2 = a^2 - b^2. This geometric relationship ensures that aa is always the largest value, representing the hypotenuse if one were to draw a right triangle from the center to a focus and then to an endpoint of the minor axis.

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Eccentricity (ee) is a numerical value that measures the 'flatness' of the ellipse, calculated as the ratio e=cae = \frac{c}{a}. Since c<ac < a, the eccentricity of an ellipse is always between 00 and 11. A value of ee near 00 results in a shape nearly identical to a circle, while a value near 11 results in a very elongated, thin shape.

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The Latus Rectum is a line segment perpendicular to the major axis that passes through one of the foci and has its endpoints on the ellipse. Visually, it indicates the 'width' of the ellipse at the focal points. Every ellipse has two such segments of equal length, calculated using the formula 2b2a\frac{2b^2}{a}.

📐Formulae

Standard Equation (Horizontal Ellipse): x2a2+y2b2=1,a>b\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b

Standard Equation (Vertical Ellipse): x2b2+y2a2=1,a>b\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, a > b

Length of Major Axis: 2a2a

Length of Minor Axis: 2b2b

Distance between Foci: 2c2c

Relation between a,b,a, b, and cc: c=a2−b2c = \sqrt{a^2 - b^2}

Eccentricity: e=cae = \frac{c}{a}

Length of Latus Rectum: L=2b2aL = \frac{2b^2}{a}

Foci (Horizontal): (±c,0)(\pm c, 0); Foci (Vertical): (0,±c)(0, \pm c)

Vertices (Horizontal): (±a,0)(\pm a, 0); Vertices (Vertical): (0,±a)(0, \pm a)

💡Examples

Problem 1:

Find the coordinates of the foci, the vertices, the lengths of the major and minor axes, the eccentricity, and the length of the latus rectum for the ellipse given by the equation x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1.

Solution:

  1. Identify orientation: Since 25>925 > 9, the denominator under x2x^2 is larger (a2=25a^2=25), so this is a horizontal ellipse.
  2. Find aa and bb: a=25=5a = \sqrt{25} = 5 and b=9=3b = \sqrt{9} = 3.
  3. Calculate cc: c=a2−b2=25−9=16=4c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4.
  4. Vertices: (±a,0)=(±5,0)(\pm a, 0) = (\pm 5, 0).
  5. Foci: (±c,0)=(±4,0)(\pm c, 0) = (\pm 4, 0).
  6. Axes Lengths: Major Axis =2a=10= 2a = 10; Minor Axis =2b=6= 2b = 6.
  7. Eccentricity: e=ca=45=0.8e = \frac{c}{a} = \frac{4}{5} = 0.8.
  8. Latus Rectum: 2b2a=2(9)5=185=3.6\frac{2b^2}{a} = \frac{2(9)}{5} = \frac{18}{5} = 3.6.

Explanation:

This approach involves comparing the given equation to the standard form x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 to extract aa and bb, then deriving all other geometric properties using standard formulae.

Problem 2:

Find the equation of the ellipse whose vertices are (0,±13)(0, \pm 13) and whose foci are (0,±5)(0, \pm 5).

Solution:

  1. Determine Orientation: Since the vertices and foci lie on the yy-axis, the ellipse is vertical.
  2. Identify aa and cc: From the coordinates, a=13a = 13 and c=5c = 5.
  3. Find b2b^2: Use the relation b2=a2−c2b^2 = a^2 - c^2. b2=132−52=169−25=144b^2 = 13^2 - 5^2 = 169 - 25 = 144.
  4. Write Equation: For a vertical ellipse, the equation is x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1. Substituting the values: x2144+y2169=1\frac{x^2}{144} + \frac{y^2}{169} = 1.

Explanation:

Because the non-zero coordinates are in the yy-position, we know the major axis is vertical. We identify the semi-major axis aa from the vertices and the focal distance cc from the foci, then solve for the semi-minor axis bb to complete the standard equation.