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Conic Sections - Standard equations of an ellipse

Grade 11CBSE

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

🔑Concepts

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An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points (foci) in the plane is a constant (2a2a).

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The constant distance 2a2a is greater than the distance between the two foci 2c2c.

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Major Axis: The line segment through the foci of the ellipse, with length 2a2a.

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Minor Axis: The line segment perpendicular to the major axis through the center, with length 2b2b.

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Eccentricity (ee): The ratio of the distance from the center of the ellipse to one of the foci to the distance from the center to one of the vertices (e=cae = \frac{c}{a}). Since c<ac < a, for an ellipse, 0<e<10 < e < 1.

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Latus Rectum: The line segment perpendicular to the major axis through any of the foci and whose endpoints lie on the ellipse. Its length is 2b2a\frac{2b^2}{a}.

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Relationship between semi-major axis (aa), semi-minor axis (bb), and distance of focus from center (cc): c2=a2−b2c^2 = a^2 - b^2.

📐Formulae

x2a2+y2b2=1,(a>b) [Horizontal Ellipse]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b) \text{ [Horizontal Ellipse]}

x2b2+y2a2=1,(a>b) [Vertical Ellipse]\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, (a > b) \text{ [Vertical Ellipse]}

c=a2−b2c = \sqrt{a^2 - b^2}

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

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

Length of Major Axis=2a\text{Length of Major Axis} = 2a

Length of Minor Axis=2b\text{Length of Minor Axis} = 2b

💡Examples

Problem 1:

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

Solution:

Comparing the given equation with x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, we get a2=25a^2 = 25 and b2=9b^2 = 9. This implies a=5a = 5 and b=3b = 3. Since a>ba > b, the major axis is along the xx-axis.

  1. c=a2−b2=25−9=16=4c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4.
  2. Foci: (±c,0)=(±4,0)(\pm c, 0) = (\pm 4, 0).
  3. Vertices: (±a,0)=(±5,0)(\pm a, 0) = (\pm 5, 0).
  4. Length of major axis: 2a=2(5)=102a = 2(5) = 10.
  5. Length of minor axis: 2b=2(3)=62b = 2(3) = 6.
  6. Eccentricity: e=ca=45=0.8e = \frac{c}{a} = \frac{4}{5} = 0.8.
  7. Length of latus rectum: 2b2a=2(9)5=185=3.6\frac{2b^2}{a} = \frac{2(9)}{5} = \frac{18}{5} = 3.6.

Explanation:

First identify aa and bb to determine the orientation (horizontal vs vertical). Then calculate cc using c2=a2−b2c^2 = a^2 - b^2 to find the foci and eccentricity.

Problem 2:

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

Solution:

The vertices are on the yy-axis, so the ellipse is vertical with the standard form x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1. Vertices are (0,±a)=(0,±13)⇒a=13(0, \pm a) = (0, \pm 13) \Rightarrow a = 13. Foci are (0,±c)=(0,±5)⇒c=5(0, \pm c) = (0, \pm 5) \Rightarrow c = 5. We know c2=a2−b2c^2 = a^2 - b^2, so: b2=a2−c2b^2 = a^2 - c^2 b2=132−52b^2 = 13^2 - 5^2 b2=169−25b^2 = 169 - 25 b2=144b^2 = 144 Substituting a2=169a^2 = 169 and b2=144b^2 = 144 into the standard equation: x2144+y2169=1\frac{x^2}{144} + \frac{y^2}{169} = 1

Explanation:

Since the vertices and foci lie on the yy-axis, the major axis is vertical. We use the coordinates to find aa and cc, then solve for bb to construct the final equation.