Let P be a point on the ellipse x²/a² + y²/b² = 1, 0

Q: Wait, why do we use the eccentric angle $\theta$ instead of just using $x$ and $y$ directly?

That is a fantastic question. While you *could* solve this using Cartesian coordinates, you would quickly find yourself drowning in square roots and messy radical expressions. By using the eccentric angle $\theta$, we map the ellipse to a circle, which is a much friendlier geometric space. It transforms a problem of 'algebraic manipulation' into a problem of 'trigonometric identity,' which is almost always the intended path in JEE Advanced problems.

Q: Does the ratio $r:s$ actually change the shape of the locus, or is it still an ellipse?

It remains an ellipse! Think of the ratio $r:s$ as a scaling factor. The original ellipse is defined by the relationship between $a$ and $b$. By introducing the ratio $r:s$, you are essentially performing a linear transformation on the y-coordinates. Since linear transformations preserve the 'elliptical' nature of the curve, the resulting locus is simply a scaled version of the original ellipse.

Q: I got stuck at the very end. How do I know if my final equation is correct?

Check the boundary conditions. If $r=1$ and $s=0$, point $R$ should coincide with $Q$, and your equation should reduce to the auxiliary circle $x^2 + y^2 = a^2$. If $r=0$ and $s=1$, point $R$ should coincide with $P$, and your equation should reduce to the original ellipse. If your derived equation satisfies these limits, you can be confident in your result.

Question

Let $P$ be a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, 0 < b < a$. Let the line parallel to y-axis passing through $P$ meet the circle $x^2 + y^2 = a^2$ at the point $Q$ such that $P$ and $Q$ are on the same side of x-axis. For two positive real numbers $r$ and $s$, find the locus of the point $R$ on $PQ$ such that $PR : RQ = r : s$ as $P$ varies over the ellipse.

Sigma Percentile · Accepted Answer

### The Geometry of Motion: Unveiling the Locus

Welcome, fellow traveler in the world of mathematics. Today, we are going to dissect a problem that, at first glance, might seem like a dry exercise in coordinate geometry.

But I want you to pause and look at the problem statement again. We have an ellipse, a circle, a vertical line, and a point $R$ dancing between them.

This is not just algebra; this is the study of how shapes transform and relate to one another. Let us embark on this journey together.

### Phase 1: The Auxiliary Circle and the Eccentric Angle

Every ellipse, defined by the equation

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

has a secret companion: the auxiliary circle, $x^2 + y^2 = a^2$. Imagine the ellipse as a circle that has been squashed along the y-axis.

This relationship is the key to our entire solution. When we define point $P$ on the ellipse, we could use standard Cartesian coordinates $(x, y)$, but that would lead us into a labyrinth of square roots.

Instead, we use the eccentric angle $	heta$. By setting $P = (a \cos 	heta, b \sin 	heta)$, we are essentially saying that $P$ is a projection of a point on the auxiliary circle.

This is the first step in mastering JEE-level geometry: choosing the right coordinate system to make the math work for you, not against you.

### Phase 2: The Vertical Constraint

Now, consider the line $PQ$. The problem tells us it is parallel to the y-axis. This is a gift!

A vertical line means that the x-coordinate is constant for every point on that line. If $P$ has an x-coordinate of $a \cos 	heta$, then $Q$ must also have an x-coordinate of $a \cos 	heta$.

Since $Q$ lies on the auxiliary circle $x^2 + y^2 = a^2$, we substitute our x-value: $(a \cos 	heta)^2 + y_Q^2 = a^2$. Solving for $y_Q$, we find $y_Q = a \sin 	heta$.

We now have the coordinates of both points: $P = (a \cos 	heta, b \sin 	heta)$ and $Q = (a \cos 	heta, a \sin 	heta)$. Notice how the x-coordinates are identical, while the y-coordinates capture the difference between the ellipse and the circle.

### Phase 3: The Section Formula

We introduce point $R(h, k)$, which divides the segment $PQ$ in the ratio $r:s$. This is where the physics of the problem meets the algebra.

The section formula is our bridge. For the x-coordinate $h$, we have

$$h = \frac{r(a \cos 	heta) + s(a \cos 	heta)}{r+s}$$

Look at that! Because the x-coordinates of $P$ and $Q$ are the same, the ratio doesn't even matter for the x-component. It simplifies beautifully to $h = a \cos 	heta$.

Now, for the y-coordinate $k$, we apply the formula with more care:

$$k = \frac{r(a \sin 	heta) + s(b \sin 	heta)}{r+s}$$

Here, the ratio matters deeply. We factor out the $\sin 	heta$ to get

$$k = \frac{(ar + bs) \sin 	heta}{r+s}$$

This is the heart of the problem. We have successfully expressed the coordinates of $R$ in terms of the parameter $	heta$.

### Phase 4: The Final Elimination

We are almost there. We have $h = a \cos 	heta$ and $k = \frac{(ar + bs) \sin 	heta}{r+s}$. To find the locus, we need to eliminate $	heta$.

We isolate the trigonometric functions: $\cos 	heta = \frac{h}{a}$ and $\sin 	heta = \frac{k(r+s)}{ar+bs}$. Now, we invoke the most powerful identity in trigonometry: $\cos^2 	heta + \sin^2 	heta = 1$.

Substituting our expressions, we get:

$$\left(\frac{h}{a}ight)^2 + \left(\frac{k(r+s)}{ar+bs}ight)^2 = 1$$

Replacing $h$ and $k$ with the general variables $x$ and $y$, we arrive at our final equation:

$$\frac{x^2}{a^2} + \frac{y^2(r+s)^2}{(bs+ar)^2} = 1$$

### Conclusion: The Elegance of the Result

Look at the final equation. It is still an ellipse! The ratio $r:s$ has effectively modified the semi-minor axis of the original ellipse.

This is the beauty of coordinate geometry: even as we move points and change ratios, the fundamental structure—the elliptical nature of the path—remains invariant.

You have successfully navigated the geometry, applied the section formula, and used trigonometric identities to reveal the hidden path. Take a moment to appreciate this. You didn't just solve for an equation; you uncovered a geometric truth. Keep this mindset, and no problem will ever be too daunting.

Visualized Solution

Visualizing the Ellipse and Circle

Parametric Coordinates of $P$

Defining the Vertical Line $P Q$

Finding the Coordinates of $Q$

Defining the Point $R (h, k)$

Applying Section Formula for $h$

Applying Section Formula for $k$

Simplifying the Expression for $k$

Isolating $cos θ$ and $sin θ$

Eliminating the Parameter $θ$

The Final Locus Equation

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

The Geometry of Motion

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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Visualized Solution

Visualizing the Ellipse and Circle

Parametric Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Defining the Vertical Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Finding the Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Defining the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R(h,k)

Applying Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Applying Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Simplifying the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosθ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ

Eliminating the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ