Let PQ be a focal chord of the parabola y² = 4x such that it subtends an angle of π/2 at the point (

Q: Wait, why is the chord vertical? Can't it be slanted?

That is a fantastic question! When we solve the perpendicularity condition $m_{PR} \cdot m_{QR} = -1$, we arrive at the equation $(t^2 - 1)^2 = 0$. This forces $t^2 = 1$. If $t^2 = 1$, then the x-coordinate of $P$ is $1^2 = 1$, and the x-coordinate of $Q$ is $1/1^2 = 1$. Since both points have the same x-coordinate, the chord must be a vertical line, $x = 1$.

Q: Why did we use $ae = 1$ for the ellipse?

Remember that the focus of a standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is at $(ae, 0)$. The problem states that $PQ$ is a focal chord of this ellipse. Since we already established that $PQ$ is the vertical line $x = 1$, the focus of the ellipse must lie on this line. Therefore, the x-coordinate of the focus, which is $ae$, must equal $1$.

Q: Why do we reject the negative root for $a$?

In our quadratic equation $a^2 - 2a - 1 = 0$, we found $a = 1 \pm \sqrt{2}$. Since $a$ represents the semi-major axis of an ellipse, it must be a positive length. Because $\sqrt{2} \approx 1.414$, the value $1 - \sqrt{2}$ is negative. Thus, we must reject it and accept only $a = 1 + \sqrt{2}$.

Question

Let $PQ$ be a focal chord of the parabola $y^2 = 4x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point $(3, 0)$. Let the line segment $PQ$ be also a focal chord of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a^2 > b^2$. If $e$ is the eccentricity of the ellipse $E$, then the value of $\frac{1}{e^2}$ is equal to :

Options:

(A) $1 + \sqrt{2}$

(B) $3 + 2\sqrt{2}$

(C) $1 + 2\sqrt{3}$

(D) $4 + 5\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Conics: A Journey Through Focal Chords

Welcome, my dear student. Today, we are going to peel back the layers of a beautiful problem that bridges the gap between the parabola and the ellipse. It is not just about solving for a variable; it is about recognizing the hidden geometric symmetries that make conic sections so elegant.

### Phase 1: The Parabola's Secret

We begin with the parabola $y^2 = 4x$. By comparing this with the standard form $y^2 = 4ax$, we immediately identify $a = 1$. This tells us that the focus $S$ is located at $(1, 0)$.

Now, imagine a focal chord $PQ$ passing through this focus. To handle this analytically, we use parametric coordinates. If we define point $P$ as $(t^2, 2t)$, then the property of a focal chord dictates that the other end, $Q$, must have the parameter $-\frac{1}{t}$.

Thus, $Q$ is $(\frac{1}{t^2}, -\frac{2}{t})$. This is our starting point.

### Phase 2: The Right Angle Trap

Next, we introduce the point $R(3, 0)$ on the x-axis. The problem tells us that the chord $PQ$ subtends an angle of $\frac{\pi}{2}$ at $R$. This is our golden key.

In coordinate geometry, when two lines are perpendicular, the product of their slopes is $-1$. So, we calculate the slope of $PR$ and the slope of $QR$.

The slope of $PR$ is:
$$m_{PR} = \frac{2t - 0}{t^2 - 3} = \frac{2t}{t^2 - 3}$$

Similarly, the slope of $QR$ is:
$$m_{QR} = \frac{-\frac{2}{t} - 0}{\frac{1}{t^2} - 3} = \frac{-2t}{1 - 3t^2}$$

### Phase 3: The Geometric Revelation

Now, we apply the condition $m_{PR} \cdot m_{QR} = -1$. Substituting our expressions, we get:

$$\left(\frac{2t}{t^2 - 3}\right) \cdot \left(\frac{-2t}{1 - 3t^2}\right) = -1$$

This simplifies to:

$$\frac{-4t^2}{(t^2 - 3)(1 - 3t^2)} = -1$$

Canceling the negatives and cross-multiplying, we arrive at $4t^2 = (t^2 - 3)(1 - 3t^2)$. Expanding this, we get $4t^2 = t^2 - 3t^4 - 3 + 9t^2$, which rearranges to $3t^4 - 6t^2 + 3 = 0$.

Dividing by $3$, we find $(t^2 - 1)^2 = 0$, which means $t^2 = 1$. This is the moment of clarity! If $t^2 = 1$, then the x-coordinates of $P$ and $Q$ are both $1$. Our chord is the vertical line $x = 1$, which is the latus rectum of the parabola.

### Phase 4: The Ellipse Connection

Now, we transition to the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We are told $PQ$ is also a focal chord of this ellipse. Since $PQ$ is the vertical line $x = 1$ and it passes through the focus $(ae, 0)$, we must have $ae = 1$.

Furthermore, because $PQ$ is a vertical focal chord, it is the latus rectum of the ellipse. The length of $PQ$ is the distance from $y = 2$ to $y = -2$, which is $4$.

The formula for the length of the latus rectum is $\frac{2b^2}{a}$. Equating these, we get:

$$\frac{2b^2}{a} = 4 \Rightarrow b^2 = 2a$$

### Phase 5: The Final Calculation

We use the fundamental identity $b^2 = a^2(1 - e^2) = a^2 - (ae)^2$. Substituting $b^2 = 2a$ and $ae = 1$, we get:

$$2a = a^2 - 1 \Rightarrow a^2 - 2a - 1 = 0$$

Solving this quadratic, we find $a = 1 + \sqrt{2}$ (rejecting the negative root). Finally, we need $\frac{1}{e^2}$. Since $ae = 1$, we have $e = \frac{1}{a}$, so $\frac{1}{e^2} = a^2$.

Squaring $1 + \sqrt{2}$, we get $1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}$.

And there you have it! By trusting the geometry, we turned a daunting algebraic problem into a beautiful, logical progression. The final answer is **$3 + 2\sqrt{2}$**.

Select Answer:

Visualized Solution

Visualizing the Parabola

Parametric Coordinates of Focal Chord

The Orthogonality Condition

Calculating the Slopes

Applying the Perpendicularity Condition

Solving for Parameter $t$

Identifying the Exact Focal Chord

Transition to the Ellipse

Latus Rectum of the Ellipse

The Eccentricity Relation

Solving for Semi-Major Axis $a$

Calculating $\frac{1}{e ^{2}}$

Final Conclusion

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

The Geometry of Conics

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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