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 option (A) \(4x^2 + y^2 = 4\).
1Step 1: Determine the Semi-minor Axis
The equation \((x-1)^2 + y^2 = 1\) is a circle with center \((1, 0)\) and radius \(1\). A diameter of this circle is \(2\). Since a diameter is double the radius, the radius itself is \(1\). The semi-minor axis of the ellipse will be half of this diameter, which is \(1\).
2Step 2: Determine the Semi-major Axis
The equation \(x^2 + (y-2)^2 = 4\) represents a circle with center \((0, 2)\) and radius \(2\). A diameter of this circle is \(4\). The semi-major axis of the ellipse will be half of this diameter, which is \(2\).
3Step 3: Determine the Equation of the Ellipse
The standard form of the ellipse with semi-major axis \(a\) along the y-axis and semi-minor axis \(b\) along the x-axis, centered at the origin, is given by:\[\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\]where \(a = 2\) and \(b = 1\). Substituting these values we get:\[\frac{x^2}{1^2} + \frac{y^2}{2^2} = 1\]Simplifying, the equation becomes:\[x^2 + \frac{y^2}{4} = 1\]or equivalently:\[4x^2 + y^2 = 4\]
4Step 4: Match with Given Options
The equation we derived is \(4x^2 + y^2 = 4\), which matches option (A) \(4x^2 + y^2 = 4\). Therefore, the correct option is (A).
Key Concepts
Semi-Major AxisSemi-Minor AxisCoordinates of Center
Semi-Major Axis
The semi-major axis is a key component in defining the shape of an ellipse. It is the longest radius that runs from the center to the edge, touching the ellipse's farthest point from the center.
In the original problem, the semi-major axis is derived from the circle with the equation \( x^2 + (y-2)^2 = 4 \). This equation represents a circle centered at \((0, 2)\) with a radius of \(2\). The diameter of this circle is \(4\), meaning it extends \(2\) units in either direction from the center on the y-axis.
In the original problem, the semi-major axis is derived from the circle with the equation \( x^2 + (y-2)^2 = 4 \). This equation represents a circle centered at \((0, 2)\) with a radius of \(2\). The diameter of this circle is \(4\), meaning it extends \(2\) units in either direction from the center on the y-axis.
- The length of the semi-major axis is thus half the diameter, which is \(2\).
- This axis runs vertically along the y-axis, meaning it directly influences how far the ellipse stretches upwards and downwards from its center.
Semi-Minor Axis
Similar to the semi-major axis, the semi-minor axis defines the ellipse, but it represents the shortest radius from the center to the ellipse's edge.
This problem uses the circle \((x-1)^2 + y^2 = 1\) to define the semi-minor axis. This circle has a center at \((1, 0)\) and a radius of \(1\). Its diameter stretches \(2\) units across, meaning the semi-minor axis is half this distance.
This problem uses the circle \((x-1)^2 + y^2 = 1\) to define the semi-minor axis. This circle has a center at \((1, 0)\) and a radius of \(1\). Its diameter stretches \(2\) units across, meaning the semi-minor axis is half this distance.
- The semi-minor axis therefore measures \(1\).
- It extends horizontally, aligning with the x-axis and dictating the ellipse's width.
Coordinates of Center
The coordinates of the center describe the exact point from which both the semi-major and semi-minor axes extend.
In this exercise, it's provided that the center of the ellipse is at the origin, which is the point \((0, 0)\). This position is crucial because it serves as a reference for plotting both axes and therefore the entire ellipse.
In this exercise, it's provided that the center of the ellipse is at the origin, which is the point \((0, 0)\). This position is crucial because it serves as a reference for plotting both axes and therefore the entire ellipse.
- Since an ellipse is symmetrically centered around this point, the center at \((0, 0)\) ensures the ellipse is balanced on both the x and y axes.
- Knowing the center helps in deriving the standard form equation of the ellipse, ensuring precise calculations and graphing.
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 \(y^{2}\) \(=4 x\) are at right angles, then the locus of the point \(P\) is (A) \(2 x+1=0\) (B) \(x=-1
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