An ellipse has eccentricity 1/2 and one focus at the point P(1/2, 1). Its one directrix is the commo

Q: Wait, why did we choose $x = 1$ as the directrix and not $x = -1$?

Great observation! The problem explicitly states the directrix is the one *nearer* to the focus $P\left(\frac{1}{2}, 1\right)$. The distance from $P$ to $x = 1$ is $|\frac{1}{2} - 1| = \frac{1}{2}$, while the distance to $x = -1$ is $|\frac{1}{2} - (-1)| = \frac{3}{2}$. We always follow the constraints given in the problem statement to define our geometry.

Q: Why do we square the equation $PS = e \cdot PM$?

It is a strategic choice to avoid the square root function. $PS = \sqrt{(x-x_0)^2 + (y-y_0)^2}$. If we keep the root, the algebra becomes a nightmare. Squaring both sides, $PS^2 = e^2 \cdot PM^2$, keeps the equation polynomial and much easier to handle. It is a classic JEE trick to keep your workspace clean and error-free.

Q: How do I know if I should complete the square or leave it in general form?

The question asks for the 'standard form' of the ellipse. This implies the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. Completing the square is the only way to shift the center from the origin to $(h, k)$ and identify the semi-major and semi-minor axes. It is the final step in revealing the true identity of the conic.

Question

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left(\frac{1}{2}, 1\right)$. 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.........

Sigma Percentile · Accepted Answer

### The Geometry of the Dance: Unveiling the Ellipse

Welcome, future engineer. Today, we are not just solving an equation; we are uncovering a hidden structure.

Imagine you are standing in a coordinate plane. You see a circle, $x^2 + y^2 = 1$, perfectly centered at the origin, and a hyperbola, $x^2 - y^2 = 1$, standing tall. Between them, there is a common tangent—a line that touches both curves simultaneously.

### Phase 1: The Hunt for the Directrix

Before we can define our ellipse, we need its directrix. Look closely at the circle and the hyperbola. The circle has a radius of $1$, and the hyperbola's vertices are at $(\pm 1, 0)$.

It is visually intuitive that the vertical lines $x = 1$ and $x = -1$ are the common tangents. The problem provides a focus $P\left(\frac{1}{2}, 1\right)$ and specifies that the directrix is the one *nearer* to this point.

A quick calculation shows the distance to $x = 1$ is $\frac{1}{2}$, while the distance to $x = -1$ is $\frac{3}{2}$. The choice is clear: our directrix is the line $x = 1$.

### Phase 2: The Conic Definition

Now, we invoke the golden rule of conic sections. For any point $(x, y)$ on our ellipse, the distance to the focus $S$, denoted as $PS$, must be equal to the eccentricity $e$ multiplied by the perpendicular distance to the directrix, $PM$.

That is, $PS = e \cdot PM$. We are given $e = \frac{1}{2}$. To avoid the terror of square roots, we square both sides: $PS^2 = e^2 \cdot PM^2$.

Substituting our values, we get:

$$ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = \left(\frac{1}{2}\right)^2 (x - 1)^2 $$

### Phase 3: The Algebraic Grind

Let's expand carefully. The left side becomes $\left(x^2 - x + \frac{1}{4}\right) + (y^2 - 2y + 1)$. The right side is $\frac{1}{4}(x^2 - 2x + 1)$.

To clear the fraction, we multiply the entire equation by $4$. This transforms our equation into:

$$ 4x^2 - 4x + 1 + 4y^2 - 8y + 4 = x^2 - 2x + 1 $$

By gathering all terms on one side, we arrive at the general form:

$$ 3x^2 - 2x + 4y^2 - 8y + 4 = 0 $$

### Phase 4: The Final Reveal

We are in the home stretch. To reach the standard form, we must complete the square. We group the $x$ terms and the $y$ terms:

$$ 3\left(x^2 - \frac{2}{3}x\right) + 4(y^2 - 2y) + 4 = 0 $$

By adding and subtracting the necessary constants, we transform this into:

$$ 3\left(x - \frac{1}{3}\right)^2 + 4(y - 1)^2 = \frac{1}{3} $$

Finally, dividing by $\frac{1}{3}$, we obtain the elegant standard form:

$$ \frac{\left(x - \frac{1}{3}\right)^2}{\left(\frac{1}{3}\right)^2} + \frac{(y - 1)^2}{\left(\frac{1}{2\sqrt{3}}\right)^2} = 1 $$

Look at that result. From a simple set of curves and a focus, we have derived the precise equation of an ellipse. This is the power of coordinate geometry—turning abstract relationships into tangible, beautiful equations. You have done well.

Visualized Solution

Given Curves & Focus

Common Tangents

Identifying the Directrix

Conic Section Definition

Substituting the Values

Expanding the Equation

Rearranging Terms

Completing the Square

Standard Form of the Ellipse

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

The Geometry of the Dance

Phase 1

Phase 2

Phase 3

Phase 4

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