Let P be a point on the ellipse x²/9 + y²/4 = 1. Let the line passing through P and parallel to y-ax

Q: Wait, why do we use $(3\cos\theta, 2\sin\theta)$ for $P$ instead of just $(x, y)$?

Great question! While you *could* use $(x, y)$, it would lead to a messy algebraic nightmare. By using the parametric form $(a\cos\theta, b\sin\theta)$, we reduce the entire problem to a single variable $\theta$. This makes the elimination process much cleaner and helps us identify the conic section immediately.

Q: Why is the $x$-coordinate of $Q$ the same as $P$?

The problem explicitly states that the line passing through $P$ is parallel to the $y$-axis. A line parallel to the $y$-axis is a vertical line, meaning every point on that line has the same $x$-coordinate. Since $Q$ lies on this vertical line, $x_Q$ must equal $x_P$.

Q: How do I know if the major axis is along the $x$-axis or $y$-axis in the final locus?

Look at the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. If $A^2 > B^2$, the major axis is along the $x$-axis. If $B^2 > A^2$, it's along the $y$-axis. In our case, we compared $9$ and $\frac{324}{49}$, and since $9 > \frac{324}{49}$, the major axis is horizontal.

Question

Let $P$ be a point on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Let the line passing through $P$ and parallel to $y$-axis meet the circle $x^2 + y^2 = 9$ at point $Q$ such that $P$ and $Q$ are on the same side of the $x$-axis. Then, the eccentricity of the locus of the point $R$ on $PQ$ such that $PR : RQ = 4 : 3$ as $P$ moves on the ellipse, is :

Options:

(A) $\frac{11}{19}$

(B) $\frac{13}{21}$

(C) $\frac{\sqrt{139}}{23}$

(D) $\frac{\sqrt{13}}{7}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, watching a point $P$ trace the path of an ellipse defined by:

$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

This is a symmetric shape with a semi-major axis $a=3$ and a semi-minor axis $b=2$.

Now, consider the circle $x^2 + y^2 = 9$. Since the radius of this circle is $3$, it matches the semi-major axis of our ellipse, making it the **auxiliary circle** of the ellipse.

### The Vertical Bridge

To track point $P$, we utilize parametric coordinates. We define $P$ as $(3\cos	heta, 2\sin	heta)$.

The problem introduces a vertical line passing through $P$. Because this line is parallel to the $y$-axis, the $x$-coordinate of any point on it is constant.

Thus, the point $Q$ on the circle must share the same $x$-coordinate as $P$: $x_Q = 3\cos	heta$. Substituting this into the circle's equation:

$$(3\cos	heta)^2 + y_Q^2 = 9 \Rightarrow y_Q^2 = 9(1 - \cos^2	heta) = 9\sin^2	heta$$

Assuming $P$ and $Q$ are on the same side of the $x$-axis, we take the positive root, yielding $Q(3\cos	heta, 3\sin	heta)$.

### The Birth of the Locus

We introduce point $R$, which divides the segment $PQ$ in the ratio $4:3$. Using the section formula, we find the coordinates $(h, k)$ of $R$.

Since the segment is vertical, the $x$-coordinate remains $h = 3\cos	heta$. For the $y$-coordinate, we apply the section formula:

$$k = \frac{4(3\sin	heta) + 3(2\sin	heta)}{4 + 3} = \frac{12\sin	heta + 6\sin	heta}{7} = \frac{18\sin	heta}{7}$$

To find the locus, we eliminate $	heta$ using the identity $\cos^2	heta + \sin^2	heta = 1$. Given $\cos	heta = \frac{h}{3}$ and $\sin	heta = \frac{7k}{18}$, we obtain:

$$\left(\frac{h}{3}ight)^2 + \left(\frac{7k}{18}ight)^2 = 1$$

Replacing $h$ and $k$ with $x$ and $y$, the equation of our new locus is:

$$\frac{x^2}{9} + \frac{49y^2}{324} = 1$$

### The Final Reveal

We rewrite the equation as:

$$\frac{x^2}{9} + \frac{y^2}{324/49} = 1$$

Comparing this to the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, we identify $A^2 = 9$ and $B^2 = \frac{324}{49}$. Since $A^2 > B^2$, the major axis is horizontal.

The eccentricity $e$ is calculated as follows:

$$e = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{324/49}{9}} = \sqrt{1 - \frac{36}{49}} = \sqrt{\frac{13}{49}}$$

The final eccentricity is:

**$e = \frac{\sqrt{13}}{7}$**

Select Answer:

Visualized Solution

Visualize the Geometry

Parametric Point $P$ on Ellipse

Finding Point $Q$ on the Circle

Calculating $y$ -coordinate of $Q$

Defining Point $R$ on $P Q$

The Section Formula

Applying the Section Formula

Simplifying $k$

Eliminating the Parameter $θ$

Finding the Locus Equation

Standard Form of Locus

Eccentricity Formula

Calculating Eccentricity

Final Conclusion

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

Analyzing the Setup

The Vertical Bridge

The Birth of the Locus

The Final Reveal

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