This problem requires us to find the angle formed at the origin by the two lines connecting the origin to the intersection points of an ellipse and a straight line. This is a classic application of the homogenization technique in coordinate geometry, followed by the formula for the angle between a pair of straight lines.

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1. Key Concepts and Formulas

1.1. Understanding the Problem We are given:

• An equation of a second-degree curve (an ellipse): $x^2 + 2y^2 = 2$.
• An equation of a straight line: $x+y=1$.

The line intersects the curve at two points, say P and Q. Our goal is to find the angle $\theta$ subtended by the line segment PQ at the origin (O). This means we need to find the angle between the lines OP and OQ.

1.2. Homogenization Technique The most efficient way to find the combined equation of the pair of straight lines OP and OQ is through homogenization.

• What is Homogenization? It's a powerful technique used to derive the combined equation of two lines that pass through the origin and also pass through the intersection points of a given curve and a straight line. We achieve this by manipulating the curve's equation to make all its terms of the same degree (typically degree 2 for conic sections), using the equation of the given straight line.

• Why use it? When a pair of lines passes through the origin, their combined equation is always homogeneous of degree 2, i.e., of the form $Ax^2 + 2Hxy + By^2 = 0$. Homogenization transforms the curve's equation into this specific form, directly giving us the desired pair of lines OP and OQ.

• How it works: If a line $L: lx+my+n=0$ (where $n \neq 0$) intersects a second-degree curve $S: Ax^2+2H'xy+By^2+2Gx+2Fy+C=0$ at points P and Q, then the equation of the pair of lines OP and OQ can be found by making the equation of the curve $S$ homogeneous of degree 2 with the help of the line $L$. From the line equation, we can write $1 = \frac{lx+my}{-n}$. We use this to convert all terms of degree less than 2 in the curve's equation into terms of degree 2:

  • A constant term $C$ is replaced by $C \cdot \left(\frac{lx+my}{-n}\right)^2$.
  • Linear terms $2Gx$ and $2Fy$ are replaced by $2Gx \cdot \left(\frac{lx+my}{-n}\right)$ and $2Fy \cdot \left(\frac{lx+my}{-n}\right)$ respectively. The resulting equation will be of the form $Ax^2 + 2H_{new}xy + B_{new}y^2 = 0$.

1.3. Angle between a Pair of Straight Lines For a pair of straight lines given by the homogeneous equation $Ax^2 + 2Hxy + By^2 = 0$, the acute angle $\theta$ between them is given by the formula: $\tan \theta = \left| \frac{2\sqrt{H^2 - AB}}{A+B} \right|$ Important Note: For the lines to be real and distinct, we must have $H^2 - AB > 0$. If $H^2 - AB = 0$, the lines are coincident. If $H^2 - AB < 0$, the lines are imaginary.

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2. Step-by-Step Solution

Step 2.1: Identify the Given Equations We are provided with:

• Equation of the curve (an ellipse): $x^2 + 2y^2 = 2 \quad \text{(Equation 1)}$
• Equation of the line: $x + y = 1 \quad \text{(Equation 2)}$ Our objective is to find the angle between the lines joining the origin to the intersection points of Equation 1 and Equation 2.

Step 2.2: Prepare for Homogenization From Equation 2, we have $x+y=1$. We will use this identity to make Equation 1 homogeneous. Notice that Equation 1 has a constant term '2' on the right side. To make the entire equation homogeneous of degree 2, we need to multiply this constant by a term of degree 2. Since $1 = x+y$, we can use $1^2 = (x+y)^2$. This substitution ensures that the constant term '2' is transformed into a term of degree 2, matching the degree of $x^2$ and $y^2$ on the left side.

Step 2.3: Homogenize the Equation of the Curve Substitute $1 = (x+y)$ into Equation 1 to transform it into a homogeneous equation of degree 2: $x^2 + 2y^2 = 2 \cdot (1)^2$ Now, substitute $1 = (x+y)$ into the right side: $x^2 + 2y^2 = 2(x+y)^2$ Next, expand the term $(x+y)^2$: $x^2 + 2y^2 = 2(x^2 + 2xy + y^2)$ Distribute the '2' on the right side: $x^2 + 2y^2 = 2x^2 + 4xy + 2y^2$ Finally, rearrange all terms to one side to get the standard form $Ax^2 + 2Hxy + By^2 = 0$: $0 = 2x^2 - x^2 + 4xy + 2y^2 - 2y^2$ $x^2 + 4xy = 0$ This resulting equation, $x^2 + 4xy = 0$, represents the combined equation of the pair of straight lines OP and OQ. These are the lines joining the origin to the points of intersection P and Q.

Step 2.4: Extract Coefficients A, H, B To use the angle formula, we need to identify the coefficients $A$, $H$, and $B$ from our homogeneous equation $x^2 + 4xy = 0$. Compare $x^2 + 4xy + 0y^2 = 0$ with the general form $Ax^2 + 2Hxy + By^2 = 0$:

• $A = 1$ (coefficient of $x^2$)
• $2H = 4 \implies H = 2$ (coefficient of $xy$. Remember $2H$, not just $H$, is the coefficient of $xy$)
• $B = 0$ (coefficient of $y^2$)

Step 2.5: Calculate the Angle using the Formula Now, substitute the values of A, H, and B into the formula for the acute angle $\theta$ between the lines: $\tan \theta = \left| \frac{2\sqrt{H^2 - AB}}{A+B} \right|$ $\tan \theta = \left| \frac{2\sqrt{(2)^2 - (1)(0)}}{1+0} \right|$ Simplify the expression inside the square root and the denominator: $\tan \theta = \left| \frac{2\sqrt{4 - 0}}{1} \right|$ $\tan \theta = \left| \frac{2\sqrt{4}}{1} \right|$ $\tan \theta = \left| \frac{2 \cdot 2}{1} \right|$ $\tan \theta = 4$ So, the angle $\theta = \tan^{-1}(4)$.

Step 2.6: Express the Angle in the Required Format We need to match our result with the given options. Recall the important inverse trigonometric identity for $x > 0$: $\tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}$ Using this identity for $x=4$: $\tan^{-1}(4) + \tan^{-1}\left(\frac{1}{4}\right) = \frac{\pi}{2}$ Rearranging the terms to express $\tan^{-1}(4)$ in the desired format: $\tan^{-1}(4) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{4}\right)$ This result matches option (A).

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3. Important Tips and Common Pitfalls

• Careful Homogenization: The most crucial step is to correctly homogenize the curve's equation. Ensure all terms in the curve's equation become of the same degree (usually 2 for conic sections) using the line equation. For example, a constant term $C$ must be multiplied by $(\text{line equation})^2$, and a linear term $Gx$ must be multiplied by $(\text{line equation})$.
• Correctly Identify A, H, B: After homogenization, meticulously identify $A$ (coefficient of $x^2$), $2H$ (coefficient of $xy$), and $B$ (coefficient of $y^2$). A very common mistake is to take $H$ as the coefficient of $xy$ instead of $2H$.
• Acute Angle: The formula $\tan \theta = \left| \frac{2\sqrt{H^2 - AB}}{A+B} \right|$ always yields the tangent of the acute angle between the lines. For problems asking for the "angle subtended", the acute angle is generally the expected answer.
• Inverse Trigonometric Identities: Be familiar with fundamental inverse trigonometric identities, such as $\tan^{-1}(x) + \tan^{-1}(1/x) = \pi/2$, as they are frequently used to match your calculated angle with the provided options.

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4. Summary and Key Takeaway

This problem beautifully illustrates the utility of the homogenization technique in coordinate geometry. By transforming the equation of the curve into a homogeneous form using the line equation, we directly obtain the combined equation of the pair of lines joining the origin to the intersection points. Once this equation is in the standard form $Ax^2 + 2Hxy + By^2 = 0$, the angle between these lines can be efficiently calculated using a standard formula. This method avoids the need to find the coordinates of the intersection points P and Q, which would be a much more tedious and error-prone approach.