An ellipse is drawn by taking a diameter of the circle (x-1)² + y² = 1 as its semi-minor axis and a

Q: Wait, why do we ignore the centers of the circles $(1,0)$ and $(0,2)$?

Great question! In this problem, the ellipse is explicitly defined to be centered at the origin $(0,0)$ with its axes aligned to the coordinate axes. The circles are merely tools to provide the lengths of the semi-axes. Their positions in the plane do not affect the dimensions of the ellipse, only their diameters do.

Q: How do I know which is the semi-major axis and which is the semi-minor axis?

By definition, the semi-major axis $a$ is the longer one, and the semi-minor axis $b$ is the shorter one. Since the diameter of the second circle is $4$ and the first is $2$, it follows that $a=4$ and $b=2$.

Q: Is it possible to confuse the radius with the diameter?

Absolutely, and that is the most common trap in this problem! Always read the question carefully. The problem states the diameter is the semi-axis, not the radius. If you used the radii ($r_1=1, r_2=2$), you would get the wrong equation. Always double-check if the problem specifies radius or diameter.

Question

An ellipse is drawn by taking 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 semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is:

Options:

(A) $4x^2 + y^2 = 4$

(B) $x^2 + 4y^2 = 8$

(C) $4x^2 + y^2 = 8$

(D) $x^2 + 4y^2 = 16$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe! Today, we are going to unravel the mystery of an ellipse. It is not just a shape; it is a beautiful dance of coordinates and constraints.

Imagine you are standing on the Cartesian plane, and you are tasked with constructing an ellipse. You are given two circles, and their properties are the keys to unlocking the equation of this ellipse. Let us embark on this journey together.

### Decoding the Red Herrings

We are given two circles: $(x-1)^2 + y^2 = 1$ and $x^2 + (y-2)^2 = 4$. A common instinct is to immediately plot these circles and obsess over their centers at $(1,0)$ and $(0,2)$.

But wait! Take a breath. The problem states the ellipse is centered at the origin $(0,0)$ and its axes are the coordinate axes.

This means the centers of the circles are irrelevant to the final equation. They are red herrings designed to test your focus. The only information that matters is the size of these circles, specifically their diameters.

### The Geometry of the Circles

Let us look at the first circle: $(x-1)^2 + y^2 = 1$. Comparing this to the standard form $(x-h)^2 + (y-k)^2 = r^2$, we see the radius $r_1 = 1$.

The problem states that a diameter of this circle is the semi-minor axis $b$. Since the diameter $d_1 = 2r_1$, we find:
$$b = 2 	imes 1 = 2$$

Now, look at the second circle: $x^2 + (y-2)^2 = 4$. Here, the radius $r_2 = \sqrt{4} = 2$.

The problem states that a diameter of this circle is the semi-major axis $a$. Thus:
$$a = 2 	imes r_2 = 2 	imes 2 = 4$$

We have successfully distilled the essence of the circles into the dimensions of our ellipse: $a=4$ and $b=2$.

### The Synthesis of the Ellipse

Now, we stand at the threshold of the final equation. We know the standard form of an ellipse centered at the origin is:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

With $a=4$ and $b=2$, we substitute these values:
$$\frac{x^2}{4^2} + \frac{y^2}{2^2} = 1$$

This simplifies to:
$$\frac{x^2}{16} + \frac{y^2}{4} = 1$$

To bring this into the elegant form, we multiply the entire equation by $16$. The first term becomes $x^2$, and the second term becomes $4y^2$.

Thus, we arrive at the final equation:
$$\mathbf{x^2 + 4y^2 = 16}$$

### The Elegance of Precision

We have navigated the traps, identified the core geometric truths, and synthesized them into a final, elegant equation. The beauty of this problem lies not just in the answer, but in the clarity of thought required to reach it.

Remember, in the world of JEE Mathematics, the most complex-looking problems often yield to the simplest, most fundamental principles. Keep practicing, keep questioning, and most importantly, keep falling in love with the logic behind the problems!

Select Answer:

Visualized Solution

Orienting the Coordinate System and Ellipse Center

Analyzing the First Circle: $(x - 1)^{2} + y^{2} = 1$

Finding the Semi-Minor Axis $b$

Analyzing the Second Circle: $x^{2} + (y - 2)^{2} = 4$

Finding the Semi-Major Axis $a$

Standard Equation of the Ellipse

Substituting $a = 4$ and $b = 2$

Simplifying to the Final Form

Final Verification and Summary

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Decoding the Red Herrings

The Geometry of the Circles

The Synthesis of the Ellipse

The Elegance of Precision

Similar Questions

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P (\frac{1}{2}, 1)$ . Its one directrix is the common tangent, nearer to the point $P$ , to the circle $x^{2} + y^{2} = 1$ and the hyperbola $x^{2} - y^{2} = 1$ . The equation of the ellipse, in the standard form, is.........

The ellipse $x^{2} + 4 y^{2} = 4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point $(4, 0)$ . Then the equation of the ellipse is:

The eccentricity of an ellipse, with its centre at the origin, is $1/2$ . If one of the directrices is $x = 4$ , then the equation of the ellipse is:

An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through which of the following points ?

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(- 3, 1)$ and has eccentricity $\frac{2}{5}$ is

The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $2$ and $22$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is

A hyperbola, having the transverse axis of length $2 sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12$ . Then its equation is