An ellipse intersects the hyperbola 2x² - 2y² = 1 orthogonally. The eccentricity of the ellipse is r

Q: Why does orthogonality imply they are confocal?

In the realm of coordinate geometry, the condition of orthogonality between two conics is a powerful constraint. It implies that the gradient of the tangent to the ellipse and the gradient of the tangent to the hyperbola at the point of intersection are negative reciprocals. This geometric condition is satisfied if and only if the two conics share the same foci. It is a beautiful, deep result in conic section theory.

Q: Can I just use $dy/dx$ to solve this?

While you could technically differentiate both equations and set the product of their slopes to $-1$, that path is a labyrinth of algebra. You would end up with a system of equations that is incredibly difficult to solve. The 'Confocal Property' is the elegant shortcut that JEE examiners want you to recognize. Always look for the geometric property before diving into the calculus.

Q: Is the eccentricity of an ellipse always less than 1?

Yes, absolutely! For an ellipse, $0 1$. This distinction is the bedrock of conic classification.

Question

An ellipse intersects the hyperbola $2x^2 - 2y^2 = 1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

Options:

(A) equation of ellipse is $x^2 + 2y^2 = 2$

(B) the foci of ellipse are $(\pm 1, 0)$

(C) equation of ellipse is $x^2 + 2y^2 = 4$

(D) the foci of ellipse are $(\pm \sqrt{2}, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are witnessing a beautiful geometric dance. We have an ellipse and a hyperbola, two distinct curves, intersecting at a perfect right angle.

This isn't a coincidence; it is a mathematical symphony. Let us break this down, step by step, and uncover the elegance hidden within the algebra.

### Decoding the Hyperbola

First, look at our hyperbola: $2x^2 - 2y^2 = 1$. It looks a bit cluttered, so let us bring it into the light by converting it to the standard form.

By dividing the entire equation by $1$, we get:

$$
\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1
$$

Now, the parameters are laid bare. We can clearly see that $a_h^2 = 1/2$ and $b_h^2 = 1/2$. This symmetry is our first clue.

### The Eccentricity Bridge

Now, we calculate the eccentricity of our hyperbola, $e_h$. The formula is $e_h = \sqrt{1 + \frac{b_h^2}{a_h^2}}$.

Substituting our values, we get:

$$
e_h = \sqrt{1 + \frac{1/2}{1/2}} = \sqrt{1 + 1} = \sqrt{2}
$$

This value, $\sqrt{2}$, is the DNA of our hyperbola. The problem states that the ellipse's eccentricity, $e_e$, is the reciprocal of this. So, $e_e = 1/\sqrt{2}$.

### The Confocal Revelation

Here is the moment of truth. The problem states the curves intersect orthogonally. In the world of JEE Advanced, this is a massive hint.

It is a fundamental theorem that if two conics intersect orthogonally, they must be **confocal**—they share the exact same foci. This is the 'Aha!' moment. We don't need to solve complex differential equations; we just need to find the foci of the hyperbola and assign them to the ellipse.

### Constructing the Ellipse

The foci of our hyperbola are given by $(\pm a_h e_h, 0)$. With $a_h = 1/\sqrt{2}$ and $e_h = \sqrt{2}$, the product $a_h e_h$ is simply $1$.

Thus, the foci are at $(\pm 1, 0)$. Since the ellipse shares these foci, its focus coordinate is $a_e e_e = 1$.

Using our known $e_e = 1/\sqrt{2}$, we find:

$$
a_e \left( \frac{1}{\sqrt{2}} \right) = 1 \Rightarrow a_e = \sqrt{2}
$$

Squaring this, we get $a_e^2 = 2$.

### The Final Assembly

Finally, we need $b_e^2$. We use the relation $b_e^2 = a_e^2(1 - e_e^2)$.

Substituting our values:

$$
b_e^2 = 2 \left( 1 - \frac{1}{2} \right) = 2 \left( \frac{1}{2} \right) = 1
$$

The standard equation of an ellipse is $\frac{x^2}{a_e^2} + \frac{y^2}{b_e^2} = 1$. Plugging in our values, we get:

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

This simplifies to the final result: **$x^2 + 2y^2 = 2$**. We have arrived at the solution by understanding the geometric soul of the problem.

Animated Solution for Mathematics - Conic Sections: An ellipse intersects the hyperbola $2 x^{2} - 2 y^{2} = 1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

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Visualized Solution

Analyze the Hyperbola $2 x^{2} - 2 y^{2} = 1$

Calculate Hyperbola Eccentricity $e_{h}$

Determine Ellipse Eccentricity $e_{e}$

The Orthogonality Condition

Find Foci of Hyperbola

Solve for Ellipse Semi-major Axis $a_{e}$

Solve for Ellipse Semi-minor Axis $b_{e}$

Construct the Ellipse Equation

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Decoding the Hyperbola

The Eccentricity Bridge

The Confocal Revelation

Constructing the Ellipse

The Final Assembly

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