Let C be the circle x² + (y - 1)² = 2, E1 and E2 be two ellipses whose centres lie at the origin and

Q: Why can't we just find the equations of the ellipses $E_1$ and $E_2$?

Because the problem doesn't give you enough information to define them! You don't need their equations. You only need the points of tangency. The geometry of the line and the midpoint condition is sufficient. In JEE Advanced, often the 'missing' information is a hint that you are over-complicating the path.

Q: How do we know the slope is $-1$?

The line is $x + y = 3$, which rearranges to $y = -x + 3$. The coefficient of $x$ is the slope, which is $-1$. This is crucial because it tells us the angle $\theta$ the line makes with the positive $x$-axis, allowing us to use the parametric form.

Q: Why is the distance $r$ negative for point $R$?

In the parametric form, $r$ represents a directed distance. If $P$ is the midpoint, $Q$ and $R$ are on opposite sides of $P$. If we define the direction towards $Q$ as positive, the direction towards $R$ must be negative. It is a beautiful way to handle vector-like displacement on a 1D line.

Question

Let $C$ be the circle $x^2 + (y - 1)^2 = 2, E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C, E_1$ and $E_2$ at $P(x_1, y_1), Q(x_2, y_2)$ and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1y_1 + x_2y_2 + x_3y_3)$ is equal to ______ .

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: A Journey Through Tangency

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the coordinate plane. We have a circle, two ellipses, and a line that dances between them.

It sounds complex, but the beauty of JEE Advanced problems lies in their hidden simplicity. Let us peel back the layers.

### Phase 1: The Anchor Point

We begin with the circle $C: x^2 + (y - 1)^2 = 2$ and the line $L: x + y = 3$. This line is our bridge, touching the circle at point $P$.

To find $P$, we substitute $y = 3 - x$ into the circle equation:
$$x^2 + (3 - x - 1)^2 = 2$$

Expanding this, we get $x^2 + (2 - x)^2 = 2$, which simplifies to:
$$x^2 + 4 + x^2 - 4x = 2 \Rightarrow 2x^2 - 4x + 2 = 0$$

Dividing by $2$, we arrive at $(x - 1)^2 = 0$. This perfect square confirms that $x_1 = 1$, and consequently, $y_1 = 2$.

Our anchor point is **$P(1, 2)$**. This is the solid ground upon which we will build the rest of our solution.

### Phase 2: The Parametric Power

Now, we face the ellipses $E_1$ and $E_2$. We do not need the equations of the ellipses; we only need the coordinates of points $Q$ and $R$.

These points lie on the line $x + y = 3$, which has a slope $m = -1$. The angle $	heta$ that the line makes with the $x$-axis satisfies $	an 	heta = -1$.

Thus, we have the directional components:
$$\cos 	heta = \frac{1}{\sqrt{2}}, \quad \sin 	heta = -\frac{1}{\sqrt{2}}$$

We use the parametric form of a line: $x = x_1 + r \cos 	heta$ and $y = y_1 + r \sin 	heta$. This allows us to 'walk' along the line from point $P$ by a distance $r$. We are given that $PQ = \frac{2\sqrt{2}}{3}$.

### Phase 3: Navigating the Line

For point $Q$, we set $r = \frac{2\sqrt{2}}{3}$. Substituting this into our parametric equations:
$$x_2 = 1 + \left( \frac{2\sqrt{2}}{3} ight) \left( \frac{1}{\sqrt{2}} ight) = 1 + \frac{2}{3} = \frac{5}{3}$$
$$y_2 = 2 + \left( \frac{2\sqrt{2}}{3} ight) \left( -\frac{1}{\sqrt{2}} ight) = 2 - \frac{2}{3} = \frac{4}{3}$$

Thus, $Q = (\frac{5}{3}, \frac{4}{3})$. Since $P$ is the midpoint of $QR$, point $R$ is at the same distance but in the opposite direction.

We set $r = -\frac{2\sqrt{2}}{3}$:
$$x_3 = 1 - \left( \frac{2\sqrt{2}}{3} ight) \left( \frac{1}{\sqrt{2}} ight) = 1 - \frac{2}{3} = \frac{1}{3}$$
$$y_3 = 2 - \left( \frac{2\sqrt{2}}{3} ight) \left( -\frac{1}{\sqrt{2}} ight) = 2 + \frac{2}{3} = \frac{8}{3}$$

Thus, $R = (\frac{1}{3}, \frac{8}{3})$.

### Phase 4: The Final Summation

We now calculate the required value: $9(x_1y_1 + x_2y_2 + x_3y_3)$.

First, we find the individual products:
$$x_1y_1 = 1 \cdot 2 = 2$$
$$x_2y_2 = \frac{5}{3} \cdot \frac{4}{3} = \frac{20}{9}$$
$$x_3y_3 = \frac{1}{3} \cdot \frac{8}{3} = \frac{8}{9}$$

Summing these values:
$$2 + \frac{20}{9} + \frac{8}{9} = 2 + \frac{28}{9}$$

Finally, multiplying by $9$:
$$9 \cdot \left( 2 + \frac{28}{9} ight) = 18 + 28 = 46$$

The final answer is **$46$**. This is the elegance of mathematics; when you trust the process, the complexity melts away.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e G eo m e t r y$

$F in d in g P o in t P (x_{1}, y_{1})$

$S o l v in g f or x_{1} an d y_{1}$

$P a r am e t r i c F or m o f t h e L in e$

$F in d in g P o in t Q (x_{2}, y_{2})$

$C a l c u l a t in g Q (x_{2}, y_{2})$

$F in d in g P o in t R (x_{3}, y_{3})$

$C a l c u l a t in g R (x_{3}, y_{3})$

$C a l c u l a t in g C oor d ina t e P r o d u c t s$

$F ina l S u mma t i o n$

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Point P(x1​,y1​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for x1​ and y1​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Point Q(x2​,y2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Q(x2​,y2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Point R(x3​,y3​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating R(x3​,y3​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Coordinate Products

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Summation