The line y = x + 1 meets the ellipse x²/4 + y²/2 = 1 at two points P and Q. If r is the radius of th

Q: Wait, why can't we just find the coordinates of $P$ and $Q$ and use the distance formula?

You absolutely could, but in the high-stakes environment of the JEE Advanced, time is your most precious resource. Solving for the roots of $3x^2 + 4x - 2 = 0$ involves the quadratic formula, which leads to messy, irrational coordinates. Using Vieta's formulas and the chord length property is not just faster; it is mathematically elegant and significantly reduces the chance of a calculation error.

Q: Why is the distance formula $d = |x_1 - x_2| \sqrt{1 + m^2}$ valid here?

Think of the segment $PQ$ as the hypotenuse of a right-angled triangle. The horizontal leg is the difference in $x$-coordinates, $|x_1 - x_2|$, and the vertical leg is the difference in $y$-coordinates, $|y_1 - y_2|$. Since the points lie on a line $y = mx + c$, we know that $|y_1 - y_2| = m|x_1 - x_2|$. By the Pythagorean theorem, the distance $d = \sqrt{|x_1 - x_2|^2 + (m|x_1 - x_2|)^2} = |x_1 - x_2| \sqrt{1 + m^2}$. It is a beautiful geometric shortcut!

Q: Do I need to find the center of the circle to solve this?

Not at all! The problem asks for the radius $r$ of a circle where $PQ$ is the diameter. Since the diameter is simply the length of the chord $PQ$, you only need to find the length of $PQ$ and divide it by two. The center of the circle is irrelevant to the calculation of its radius.

Question

The line $y = x + 1$ meets the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter then $(3r)^2$ is equal to

Options:

(A) 20

(B) 12

(C) 11

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a vast, flat coordinate plane. Before you lies an ellipse, defined by the equation:

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

It is a graceful, closed curve, a perfect loop of symmetry. Now, imagine a line, $y = x + 1$, cutting through this ellipse like a sharp blade.

Where they meet, they create two points, $P$ and $Q$. These points are the anchors of our problem. We are tasked with finding the radius of a circle that uses the segment $PQ$ as its diameter.

### The Algebraic Foundation

To find where our line and ellipse collide, we must solve their equations simultaneously. We start by simplifying the ellipse equation. Multiplying the equation by $4$ gives us:

$$x^2 + 2y^2 = 4$$

Now, we substitute the line equation $y = x + 1$ into this. This substitution is the bridge between the two shapes:

$$x^2 + 2(x + 1)^2 = 4$$

Expanding the squared term, we have $x^2 + 2(x^2 + 2x + 1) = 4$. Distributing the $2$ and combining like terms, we arrive at the quadratic equation:

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

This equation is the heartbeat of our problem. Its roots, $x_1$ and $x_2$, are the $x$-coordinates of our intersection points $P$ and $Q$.

### The Elegance of Vieta

We use Vieta's formulas to avoid the tedious quadratic formula. We know the sum of the roots is $x_1 + x_2 = -\frac{4}{3}$ and the product is $x_1x_2 = -\frac{2}{3}$.

We only need the horizontal distance between them, $|x_1 - x_2|$. Using the identity $|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1x_2}$, we substitute our values:

$$|x_1 - x_2| = \sqrt{\left(-\frac{4}{3}\right)^2 - 4\left(-\frac{2}{3}\right)} = \sqrt{\frac{16}{9} + \frac{8}{3}} = \sqrt{\frac{16 + 24}{9}} = \frac{\sqrt{40}}{3}$$

This is the horizontal span of our chord.

### The Geometric Bridge

Now, we convert this horizontal span into the actual length of the chord $PQ$. For any line with slope $m$, the distance $d$ between two points is $d = |x_1 - x_2| \sqrt{1 + m^2}$.

Our line $y = x + 1$ has a slope $m = 1$. Thus, the length of the diameter $d$ is:

$$d = \frac{\sqrt{40}}{3} \sqrt{1 + 1^2} = \frac{\sqrt{40} \cdot \sqrt{2}}{3} = \frac{\sqrt{80}}{3}$$

The radius $r$ is half of this diameter:

$$r = \frac{\sqrt{80}}{6}$$

### The Final Triumph

The question asks for the value of $(3r)^2$. Let us calculate this with precision:

$$(3r)^2 = 9r^2$$

Since $r = \frac{\sqrt{80}}{6}$, then $r^2 = \frac{80}{36}$. Substituting this back, we get:

$$9 \cdot \frac{80}{36} = \frac{80}{4} = 20$$

Through the power of algebraic manipulation and geometric insight, we have arrived at the final answer: **20**.

Animated Solution for Mathematics - Conic Sections: The line $y = x + 1$ meets the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ at two points $P$ and $Q$ . If $r$ is the radius of the circle with $P Q$ as diameter then $(3 r)^{2}$ is equal to

Select Answer:

Visualized Solution

Visualize the Geometry

Intersection and the Circle

Solving Simultaneously

Substitution

Expanding the Equation

Forming the Quadratic

Vieta's Formulas

Difference of Roots

Calculating $∣ x_{1} - x_{2} ∣$

Chord Length Formula

Calculating Diameter $P Q$

Finding the Radius $r$

Final Calculation

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

Analyzing the Setup

The Algebraic Foundation

The Elegance of Vieta

The Geometric Bridge

The Final Triumph

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