Animated Solution for Mathematics - Straight Lines: If the curve $x^2+2y^2=2$ intersects the line $x+y=1$ at two points $P$ and $Q$, then the angle subtended by the line segment $PQ$ at the origin is:

Imagine you are standing at the origin of a coordinate plane. Before you lies an ellipse, defined by the equation $x^2+2y^2=2$.

A straight line, $x+y=1$, slices through this ellipse, creating two intersection points, $P$ and $Q$. Your mission is to find the angle subtended by the line segment $PQ$ at the origin.

Many students immediately reach for their pens to solve for $P$ and $Q$ by substituting $y=1-x$ into the ellipse equation. While that works, it is a path filled with potential pitfalls—messy quadratics and tedious arithmetic.

Instead, let us embrace a more powerful, elegant tool: Homogenization.

The beauty of Homogenization lies in its ability to describe the two lines connecting the origin to the intersection points without ever needing to know the coordinates of $P$ and $Q$ themselves.

We start with the curve $x^2+2y^2=2$ and the line $x+y=1$. Notice that the line equation is already equal to $1$. We can rewrite the curve as:

Now, replace that $1$ with $(x+y)$. The equation becomes:

This is the magic of the technique: we have forced the curve equation to become homogeneous of degree two, effectively creating the combined equation of the two lines $OP$ and $OQ$.

Now, let us expand this expression. The right side, $2(x+y{)}^2$, expands to $2(x^2+y^2+2xy)$.

So, our equation is:

Subtracting $x^2$ and $2y^2$ from both sides, we are left with $0=x^2+4xy$, or simply:

This simple equation is the combined equation of the lines $OP$ and $OQ$. Comparing this to the standard homogeneous form $a{x}^2+2hxy+b{y}^2=0$, we identify $a=1$, $2h=4$ (so $h=2$), and $b=0$.

With our coefficients in hand, we use the standard formula for the acute angle $\alpha$ between two lines:

Substituting our values, we get:

Thus, the acute angle is $\alpha ={\tan }^{-1}(4)$. However, we must pause and consider the geometry. The formula gives us the acute angle, but the angle subtended by the segment $PQ$ at the origin is clearly obtuse.

Therefore, the required angle $\theta$ is $\pi -\alpha$, or $\pi -{\tan }^{-1}(4)$.

To match our result with the provided options, we use the inverse trigonometric identity ${\tan }^{-1}(x)+{\cot }^{-1}(x)=\frac{\pi }{2}$. This allows us to write ${\tan }^{-1}(4)=\frac{\pi }{2}-{\cot }^{-1}(4)$.

Substituting this back into our expression for $\theta$, we get:

Finally, using the identity ${\cot }^{-1}(x)={\tan }^{-1}(\frac{1}{x})$, we reach our destination:

The elegance of this solution lies not in the brute force of calculation, but in the structural beauty of the geometry. You have successfully navigated the trap and arrived at the correct answer.