Consider two straight lines, each of which is tangent to both the circle x² + y² = 1/2 and the parab

Q: Why do we reject $m^2 = -2$?

Because $m$ represents the slope of a real line in the Cartesian plane. Since $m^2 = -2$ yields imaginary values for $m$, those slopes cannot exist in our real coordinate system. We only care about real-valued tangents.

Q: Why is the area integral multiplied by 2?

The ellipse is symmetric about the x-axis. The region bounded between $x = 1/\sqrt{2}$ and $x = 1$ exists in both the first and fourth quadrants. Calculating the area in the first quadrant and doubling it accounts for the entire region bounded by the ellipse.

Q: Is there a faster way to find the intersection $Q$?

Yes, by symmetry! Since the circle and parabola are symmetric about the x-axis, the common tangents must also be symmetric about the x-axis. Their intersection point $Q$ must lie on the x-axis. You can often deduce this geometric property to save time.

Question

Consider two straight lines, each of which is tangent to both the circle $x^2 + y^2 = \frac{1}{2}$ and the parabola $y^2 = 4x$. Let these lines intersect at the point $Q$. Consider the ellipse whose center is at the origin $O(0, 0)$ and whose semi-major axis is $OQ$. If the length of the minor axis of this ellipse is $\sqrt{2}$, then which of the following statement(s) is (are) TRUE ?

Options:

(A) For the ellipse, the eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $1$

(B) For the ellipse, the eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $\frac{1}{2}$

(C) The area of the region bounded by the ellipse between the lines $x = \frac{1}{\sqrt{2}}$ and $x = 1$ is $\frac{1}{4\sqrt{2}}(\pi-2)$

(D) The area of the region bounded by the ellipse between the lines $x = \frac{1}{\sqrt{2}}$ and $x = 1$ is $\frac{1}{16}(\pi-2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are orchestrating a meeting between two fundamental shapes: the circle and the parabola.

Imagine standing at the origin of the Cartesian plane. To your left, a circle $x^2 + y^2 = \frac{1}{2}$ sits perfectly balanced. To your right, a parabola $y^2 = 4x$ stretches out, hungry for space.

Our quest is to find the lines that touch both—the common tangents. This is a classic JEE Advanced challenge, and it requires us to be both precise and intuitive.

### The Dance of Tangents

To find a line tangent to the parabola $y^2 = 4x$, we recall the standard form $y^2 = 4ax$, where $a=1$. The equation of any tangent to this parabola is given by:

$$y = mx + \frac{1}{m}$$

This is our 'master key'. This line must also kiss the circle $x^2 + y^2 = \frac{1}{2}$, which has a center at $(0,0)$ and a radius $r = \frac{1}{\sqrt{2}}$.

The condition for the line $mx - y + \frac{1}{m} = 0$ to be tangent to the circle is that the perpendicular distance from the center to the line must equal the radius. Using the distance formula, we set:

$$\frac{|m(0) - 0 + \frac{1}{m}|}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{2}}$$

Squaring both sides, we arrive at the polynomial:

$$\frac{1}{m^2(m^2 + 1)} = \frac{1}{2} \Rightarrow m^4 + m^2 - 2 = 0$$

Factoring this, we find $(m^2 + 2)(m^2 - 1) = 0$. Since $m$ must be real, we discard the imaginary roots and are left with $m^2 = 1$, giving us slopes of $m = \pm 1$.

Our tangents are $y = x + 1$ and $y = -x - 1$.

### The Intersection and the Birth of the Ellipse

Where do these lines meet? Solving $x + 1 = -x - 1$ yields $x = -1$, and consequently $y = 0$. Our intersection point $Q$ is $(-1, 0)$.

Now, we construct our ellipse. The center is at the origin, and the semi-major axis $a$ is the distance $OQ = 1$. We are given the minor axis length as $\sqrt{2}$, so $2b = \sqrt{2}$, meaning $b = \frac{1}{\sqrt{2}}$.

The equation of our ellipse is:

$$\frac{x^2}{1^2} + \frac{y^2}{(1/\sqrt{2})^2} = 1 \Rightarrow x^2 + 2y^2 = 1$$

This is the heart of our new shape.

### The Calculus of Area

Finally, we calculate the area bounded by the ellipse between $x = \frac{1}{\sqrt{2}}$ and $x = 1$. Because of the symmetry about the x-axis, the total area is twice the area in the first quadrant.

We integrate $y = \frac{1}{\sqrt{2}}\sqrt{1 - x^2}$ from $\frac{1}{\sqrt{2}}$ to $1$. The integral becomes:

$$Area = 2 \int_{1/\sqrt{2}}^{1} \frac{1}{\sqrt{2}}\sqrt{1 - x^2} dx = \sqrt{2} \int_{1/\sqrt{2}}^{1} \sqrt{1 - x^2} dx$$

Using the standard integral $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$, we evaluate the boundaries. At the upper limit $x=1$, the expression yields $\frac{\pi}{4}$. At the lower limit $x=\frac{1}{\sqrt{2}}$, we get $\frac{1}{4} + \frac{\pi}{8}$.

Subtracting these and multiplying by $\sqrt{2}$, we arrive at the final result:

$$\mathbf{Area = \frac{\pi - 2}{4\sqrt{2}}}$$

We have conquered the problem, proving that with patience and systematic steps, even the most complex geometry yields to logic.

Select Answer:

Visualized Solution

Visualizing the Curves

Tangent to the Parabola

Tangency Condition for Circle

Solving for Slope $m$

Equations of Common Tangents

Finding Intersection Point $Q$

Defining the Ellipse

Checking Option A

Setting up Area Integral

Applying Standard Integral

Evaluating Upper Limit

Evaluating Lower Limit

Final Area Calculation

Conclusion and Summary

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Dance of Tangents

The Intersection and the Birth of the Ellipse

The Calculus of Area

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