Problem 98
Question
An ellipse is drawn by considering a diameter of the circle \((x-1)^{2}+y^{2}=1\) as its semi-minor axis and a diameter of the circle \(x^{2}+(y-2)^{2}=4\) as its semi-major axis. If the centre of the ellipse is the origin and its axes are the coordinate axes, then the equation of the ellipse is (A) \(4 x^{2}+y^{2}=4\) (B) \(x^{2}+4 y^{2}=8\) (C) \(4 x^{2}+y^{2}=8\) (D) \(x^{2}+4 y^{2}=16\)
Step-by-Step Solution
Verified Answer
The equation of the ellipse is (A) \(4 x^{2}+y^{2}=4\).
1Step 1: Identify the properties of the first circle
The equation of the first circle is \((x-1)^2 + y^2 = 1\). Comparing it with the standard form \((x-h)^2 + (y-k)^2 = r^2\), we identify that the center of the circle is \((1, 0)\) and the radius is 1. Hence, the diameter of this circle is 2.
2Step 2: Find the semi-minor axis of the ellipse
The semi-minor axis of the ellipse is half of the diameter of the circle \((x-1)^2 + y^2 = 1\). Since the diameter is 2, the semi-minor axis is \(\frac{2}{2} = 1\).
3Step 3: Identify the properties of the second circle
The equation of the second circle is \(x^2 + (y-2)^2 = 4\). From \((x-h)^2 + (y-k)^2 = r^2\), we find the center is \((0, 2)\) and the radius is 2. Thus, the diameter of this circle is 4.
4Step 4: Find the semi-major axis of the ellipse
The semi-major axis of the ellipse is half of the diameter of the circle \(x^2 + (y-2)^2 = 4\). Since the diameter is 4, the semi-major axis is \(\frac{4}{2} = 2\).
5Step 5: Formulate the equation of the ellipse
The ellipse is centered at the origin with its semi-major axis aligned with y-axis and semi-minor axis aligned with x-axis. Thus, the equation of the ellipse is \(\frac{x^2}{1^2} + \frac{y^2}{2^2} = 1\), simplifying to \(x^2 + \frac{y^2}{4} = 1\).
6Step 6: Match the equation format to given options
Rewriting \(x^2 + \frac{y^2}{4} = 1\) as \(4x^2 + y^2 = 4\) by multiplying the entire equation by 4 gives us the equation that matches with option (A).
Key Concepts
semi-major axissemi-minor axiscoordinate geometry
semi-major axis
The semi-major axis is one of the most critical components of an ellipse. Imagine an ellipse as a stretched-out circle. The semi-major axis is the longest radius of the ellipse. It runs from the center to the furthest edge of the ellipse.
For the exercise, we see that the semi-major axis is aligned with the y-axis. It means the ellipse stretches more vertically.
For the exercise, we see that the semi-major axis is aligned with the y-axis. It means the ellipse stretches more vertically.
- The semi-major axis of our ellipse is 2, derived from half of the diameter of the circle with equation \( x^2 + (y-2)^2 = 4 \).
- This circle's diameter is 4, leading us to its semi-major axis being \( \frac{4}{2} = 2 \).
semi-minor axis
The semi-minor axis plays an equally important role in defining an ellipse's shape. It represents the shortest radius from the center to the edge, running perpendicular to the semi-major axis.For the ellipse in the given exercise, the semi-minor axis is oriented along the x-axis. This configuration means the ellipse is wider horizontally than vertically.
- The semi-minor axis measures 1, originating from the circle \((x-1)^2 + y^2 = 1\).
- Since the diameter of the circle is 2, the semi-minor axis becomes \( \frac{2}{2} = 1 \).
coordinate geometry
Coordinate geometry, or analytic geometry, involves describing shapes using coordinate systems and algebraic equations. In this branch of mathematics, ellipses can be well-explained using geometric properties and algebraic equations. Understanding ellipses in coordinate geometry includes:
- Recognizing they are defined by two key axes: the semi-major and semi-minor axes. The lengths of these axes influence the equation of the ellipse.
- Locating the center, which in this exercise is the origin \((0, 0)\).
- Understanding that the standard form of an ellipse's equation in coordinate geometry is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the semi-major and semi-minor axes.
Other exercises in this chapter
Problem 95
The ellipse \(x^{2}+4 y^{2}=4\) is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through
View solution Problem 96
If two tangents drawn from a point \(P\) to the parabola \(y_{1}^{2}\) \(=4 x\) are at right angles, then the locus of the point \(P\) is (A) \(2 x+1=0\) (B) \(
View solution Problem 99
Given: A circle, \(2 x^{2}+2 y^{2}=5\) and a parabola, \(y^{2}=4 \sqrt{5 x}\) Statement - I: An equation of a common tangent to these curves is \(y=x+\sqrt{5}\)
View solution Problem 100
The circle passing through the foci of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\) with center at \((0,3)\) has equation (A) \(x^{2}+y^{2}-6 y+7=0\) (B)
View solution