The centre of an ellipse is at (0, 0), major axis is on the y-axis. If the ellipse passes through (3, 2) and (1, 6), then what is its eccentricity ?
Understanding the Ellipse Problem
The question asks us to find the eccentricity of an ellipse given its center, the orientation of its major axis, and two points it passes through. The center is at the origin (0, 0), and the major axis lies along the y-axis. This information is crucial for setting up the correct standard equation of the ellipse.
When the major axis is on the y-axis and the center is at (0, 0), the standard equation of the ellipse is given by:
\(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)
Here, 'a' represents the semi-major axis length (along the y-axis) and 'b' represents the semi-minor axis length (along the x-axis). For the major axis to be on the y-axis, we must have \(a > b\).
Using the Given Points to Find Parameters
We are given that the ellipse passes through two points: (3, 2) and (1, 6). We can substitute the coordinates of these points into the standard equation to form a system of equations with \(a^2\) and \(b^2\).
Point 1: (3, 2)
Substitute \(x=3\) and \(y=2\) into the ellipse equation:
\(\frac{3^2}{b^2} + \frac{2^2}{a^2} = 1\)
\(\frac{9}{b^2} + \frac{4}{a^2} = 1 \quad \text{(Equation 1)}\)
Point 2: (1, 6)
Substitute \(x=1\) and \(y=6\) into the ellipse equation:
\(\frac{1^2}{b^2} + \frac{6^2}{a^2} = 1\)
\(\frac{1}{b^2} + \frac{36}{a^2} = 1 \quad \text{(Equation 2)}\)
Solving for \(a^2\) and \(b^2\)
We now have a system of two linear equations in terms of \(\frac{1}{b^2}\) and \(\frac{1}{a^2}\). We can solve this system using methods like substitution or elimination.
Let's use elimination. Multiply Equation 2 by 9:
\(9 \times \left(\frac{1}{b^2} + \frac{36}{a^2}\right) = 9 \times 1\)
\(\frac{9}{b^2} + \frac{324}{a^2} = 9 \quad \text{(Equation 3)}\)
Now, subtract Equation 1 from Equation 3:
\(\left(\frac{9}{b^2} + \frac{324}{a^2}\right) - \left(\frac{9}{b^2} + \frac{4}{a^2}\right) = 9 - 1\)
\(\frac{324}{a^2} - \frac{4}{a^2} = 8\)
\(\frac{320}{a^2} = 8\)
Solving for \(a^2\):
\(a^2 = \frac{320}{8} = 40\)
Now substitute the value of \(a^2\) back into Equation 2:
\(\frac{1}{b^2} + \frac{36}{40} = 1\)
\(\frac{1}{b^2} = 1 - \frac{36}{40}\)
Simplify the fraction \(\frac{36}{40}\): \(\frac{36}{40} = \frac{9 \times 4}{10 \times 4} = \frac{9}{10}\)
\(\frac{1}{b^2} = 1 - \frac{9}{10} = \frac{10 - 9}{10} = \frac{1}{10}\)
Solving for \(b^2\):
\(b^2 = 10\)
We have \(a^2 = 40\) and \(b^2 = 10\). Since \(40 > 10\), \(a > b\), which confirms that the major axis is indeed on the y-axis as stated in the problem.
Calculating the Eccentricity of the Ellipse
The eccentricity '\(e\)' of an ellipse with the major axis on the y-axis is given by the formula:
\(e = \sqrt{1 - \frac{b^2}{a^2}}\)
Substitute the values of \(a^2\) and \(b^2\):
\(e = \sqrt{1 - \frac{10}{40}}\)
\(e = \sqrt{1 - \frac{1}{4}}\)
\(e = \sqrt{\frac{4 - 1}{4}}\)
\(e = \sqrt{\frac{3}{4}}\)
\(e = \frac{\sqrt{3}}{\sqrt{4}}\)
\(e = \frac{\sqrt{3}}{2}\)
The eccentricity of the ellipse is \(\frac{\sqrt{3}}{2}\).
Summary of Steps
- Identify the standard equation of the ellipse based on the center and major axis orientation.
- Substitute the given points into the equation to form a system of equations for \(a^2\) and \(b^2\).
- Solve the system of equations to find the values of \(a^2\) and \(b^2\).
- Verify that the condition for the major axis (a > b) is met.
- Use the values of \(a^2\) and \(b^2\) to calculate the eccentricity using the appropriate formula.
Revision Table: Key Ellipse Concepts
| Concept | Description | Relevant Formula/Equation (Center at (0,0)) |
|---|---|---|
| Standard Equation (Major axis on x-axis) | Semi-major axis is 'a' (along x), Semi-minor axis is 'b' (along y), \(a > b\). | \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) |
| Standard Equation (Major axis on y-axis) | Semi-major axis is 'a' (along y), Semi-minor axis is 'b' (along x), \(a > b\). | \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) |
| Eccentricity (e) | A measure of how much the ellipse deviates from a circle (\(0 \le e < 1\)). | \(e = \sqrt{1 - \frac{b^2}{a^2}}\) (for major axis on x or y axis) |
| Relationship between a, b, and c | \(c\) is the distance from the center to a focus. | \(c^2 = a^2 - b^2\) |
Additional Information on Ellipse Properties
The eccentricity \(e\) is a key property of an ellipse. It tells us about the shape of the ellipse:
- If \(e\) is close to 0, the ellipse is nearly a circle.
- If \(e\) is close to 1, the ellipse is very elongated.
- A circle is a special case of an ellipse with \(e=0\) (where \(a=b\)).
In this problem, we found \(e = \frac{\sqrt{3}}{2}\). Since \(\sqrt{3} \approx 1.732\), \(\frac{\sqrt{3}}{2} \approx 0.866\). This value is between 0 and 1, as expected for an ellipse, and indicates a somewhat elongated shape.
The foci of the ellipse are located at a distance \(c = ae\) from the center along the major axis. Since the major axis is on the y-axis, the foci are at \((0, \pm c)\).
\(c = \sqrt{a^2 - b^2} = \sqrt{40 - 10} = \sqrt{30}\)
So, the distance from the center to the foci is \(\sqrt{30}\). The foci are at \((0, \pm \sqrt{30})\).
The vertices are the endpoints of the major and minor axes. For this ellipse:
- Vertices on the major axis (y-axis): \((0, \pm a) = (0, \pm \sqrt{40}) = (0, \pm 2\sqrt{10})\)
- Vertices on the minor axis (x-axis): \(( \pm b, 0) = ( \pm \sqrt{10}, 0)\)
Similar Questions
A man running round a racecourse notes that the sum of the distance of two flag-posts from him is always 10 m and the distance between the flag-posts is 8 m. The area of the path he encloses is
The sum of the focal distances of a point on an ellipse is constant and equal to the
What is the equation of the ellipse having foci (±2, 0) and the eccentricity \(\frac{1}{4}?\)
- What is the eccentricity of the ellipse if the angle between the straight lines joining the foci to an extremity of the minor axis is \(90^\circ\)?
Consider any point P on the ellipse \(\frac{{{{\rm{x}}^2}}}{{25}} + \frac{{{{\rm{y}}^2}}}{9} = 1\) in the first quadrant. Let r and s represent its distance from (4, 0) and (-4, 0) respectively, then (r + s) is equal to
Let P(x, y) be any point on the ellipse 25x 2+ 16y 2= 400. If Q(0, 3) and R(0, -3) are two points, then what is (PQ + PR) equal to?
What is PE + PF equal to ?
Consider the following points :
1. \(\left(\frac{\sqrt{3}}{2}, 0\right)\)
2. \(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}\right)\)
3. \(\left(\frac{\sqrt{3}}{2},-\frac{1}{4}\right)\)
Which of the above points lie on latus rectum of ellipse ?
What is the distance between the foci of the ellipse x 2+ 2y 2= 1 ?
The centre and one of the foci \((F)\) of an ellipse are at \((0, 0)\) and \((-c, 0)\) respectively. If \(P(x, y)\) is any point on the ellipse and \(2a\) is the length of the major axis, then what is \(PF\) equal to?
Important Questions from Ellipse
The equation of an ellipse which has a focus (6, 7), a directix x + y + 2 = 0 and eccentricity \(\frac{1}{{\sqrt 3 }}\), is:
The equation of the tangent at the point (x', y') to the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) is:
The equation \(\frac{{{x^2}}}{{2 - r}} + \frac{{{y^2}}}{{r - 6}} + 1 = 0\) represents an ellipse if
The equation of sphere is x2 + y2 + z2 - x + z - 2 = 0, its radius is
If the straight line x cosα + y sinα = p is tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). then
