The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x

Q: Why is the directrix $x = -\frac{a}{e}$? Can't it be something else?

The directrix is a geometric property defined by the ratio of distances. For an ellipse centered at the origin, the directrices are vertical lines at $x = \pm \frac{a}{e}$. Since we are given $x = -4$, we know it's the left directrix, which corresponds to the negative value.

Q: Why can't I just find the slope of the tangent and take the negative reciprocal?

You absolutely can! The normal is perpendicular to the tangent. However, using the standard normal formula $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$ is much faster and less prone to algebraic errors during differentiation. It is a powerful tool in your JEE arsenal.

Q: How do I know if the ellipse is horizontal or vertical?

The directrix $x = -4$ is a vertical line. This implies the major axis lies along the x-axis, meaning it is a horizontal ellipse where $a > b$. If the directrix were $y = k$, the ellipse would be vertical.

Question

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$. If one of its directrices is $x = -4$, then the equation of the normal to it at $(1, \frac{3}{2})$ is:

Options:

(A) $2y - x = 2$

(B) $4x - 2y = 1$

(C) $4x + 2y = 7$

(D) $x + 2y = 4$

Sigma Percentile · Accepted Answer

### The Geometry of the Ellipse: A Journey into Conic Sections

Imagine you are standing before a perfectly balanced ellipse, centered at the origin of your coordinate plane. It is a shape of elegance, defined by its eccentricity $e = \frac{1}{2}$, which tells us just how much this circle has been stretched.

Today, we are going to peel back the layers of this ellipse to find the equation of its normal at a specific point. This isn't just about plugging numbers into formulas; it's about understanding the geometric soul of the conic section.

### Phase 1: Decoding the Directrix

Every ellipse has an anchor, a geometric boundary that defines its shape: the directrix. For an ellipse centered at the origin, the directrices are vertical lines defined by the equation $x = \pm \frac{a}{e}$.

We are given that one of these directrices is $x = -4$. By setting $-\frac{a}{e} = -4$, we immediately see that $\frac{a}{e} = 4$.

Since we know the eccentricity $e = \frac{1}{2}$, we can solve for the semi-major axis $a$:
$$a = 4 	imes \frac{1}{2} = 2$$

With $a = 2$, we find that $a^2 = 4$. This value, $a^2$, is the heartbeat of our ellipse, defining its horizontal reach.

### Phase 2: The Geometry of the Ellipse

Now that we have the semi-major axis, we need the semi-minor axis, $b$. The relationship between $a$, $b$, and $e$ is one of the most beautiful identities in coordinate geometry:
$$b^2 = a^2(1 - e^2)$$

Let's substitute our known values:
$$b^2 = 4\left(1 - \left(\frac{1}{2}ight)^2ight)$$

Calculating this, we get:
$$b^2 = 4\left(1 - \frac{1}{4}ight) = 4\left(\frac{3}{4}ight) = 3$$

We have successfully defined the ellipse:
$$\frac{x^2}{4} + \frac{y^2}{3} = 1$$

It is a horizontal ellipse, slightly flattened, waiting for us to find the normal at the point $P(1, \frac{3}{2})$.

### Phase 3: The Normal Line

We are now at the final stage of our journey. We need the equation of the normal at $P(1, \frac{3}{2})$.

The standard equation of the normal to an ellipse at $(x_1, y_1)$ is given by:
$$\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$$

Substituting our values $a^2 = 4$, $b^2 = 3$, $x_1 = 1$, and $y_1 = \frac{3}{2}$, we get:
$$\frac{4x}{1} - \frac{3y}{\frac{3}{2}} = 4 - 3$$

Simplifying this, the first term becomes $4x$, and the second term becomes $3y 	imes \frac{2}{3} = 2y$. The right side is simply $1$.

Thus, we arrive at the final equation:
**$4x - 2y = 1$**

This is the line that stands perfectly perpendicular to the tangent at our point $P$. It is a simple, clean result that emerges from the complex geometry of the ellipse.

Animated Solution for Mathematics - Conic Sections: The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directrices is $x = - 4$ , then the equation of the normal to it at $(1, \frac{3}{2})$ is:

Select Answer:

Visualized Solution

Visualizing the Given Data

Directrix Formula

Finding the Semi-Major Axis $a$

Calculating $a$

Relation Between $a, b,$ and $e$

Substituting Values for $b^{2}$

Calculating $b^{2}$

Identifying the Point of Interest

Equation of the Normal

Substituting into the Normal Equation

Simplifying the Equation

Final Equation of the Normal

The Sigma Insight: Equation of Tangent and Normal

The Geometry of the Ellipse

Phase 1

Phase 2

Phase 3

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