A hyperbola having the transverse axis of length √2 has the same foci as that of the ellipse of 3x²

Q: Wait, how do I know the hyperbola is horizontal? Could it be vertical?

Great question! Look at the foci. We found them at $(\pm 1, 0)$. Since the y-coordinate is zero, the foci lie on the x-axis. By definition, the transverse axis of a hyperbola lies along the line connecting the foci. Therefore, the hyperbola must be horizontal.

Q: Why do we use $b^2 = a^2(e^2-1)$ for the hyperbola but $b^2 = a^2(1-e^2)$ for the ellipse?

It is all about the eccentricity $e$. For an ellipse, $e 1$, so $e^2-1$ is required to keep $b^2$ positive. Think of it as the 'correction factor' to ensure our geometry remains real and physical!

Q: Is there a faster way to check the points without plugging them in?

In this specific case, direct substitution is the most robust method. Since the equation $2x^2 - 2y^2 = 1$ is quite simple, plugging in the coordinates is actually faster than trying to derive a property of the hyperbola that might exclude a point. Trust the algebra!

Question

A hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse of $3x^2 + 4y^2 = 12$, then this hyperbola does not pass through which of the following points?

Options:

(A) $(-\sqrt{\frac{3}{2}}, 1)$

(B) $(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}})$

(C) $(\frac{1}{\sqrt{2}}, 0)$

(D) $(1, -\frac{1}{\sqrt{2}})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, coordinate-mapped plane. You see two shapes: an ellipse, elegant and closed, and a hyperbola, bold and infinite.

They seem different, yet they share a profound, hidden connection: they share the exact same foci. This is the key to unlocking our problem.

### Unmasking the Ellipse

We begin with the ellipse defined by $3x^2 + 4y^2 = 12$. To understand its soul, we must bring it into its standard form.

Dividing by $12$, we get:

$$\frac{x^2}{4} + \frac{y^2}{3} = 1$$

Here, we see $a_e^2 = 4$ and $b_e^2 = 3$. The eccentricity $e_e$ is the measure of how 'squashed' this ellipse is. Using the relation $b_e^2 = a_e^2(1 - e_e^2)$, we substitute our values:

$$3 = 4(1 - e_e^2)$$

A quick algebraic shuffle reveals $1 - e_e^2 = \frac{3}{4}$, meaning $e_e^2 = \frac{1}{4}$, so $e_e = \frac{1}{2}$.

The foci of an ellipse are located at $(\pm a_e e_e, 0)$. With $a_e = 2$ and $e_e = \frac{1}{2}$, our foci are at $(\pm 1, 0)$. These two points are the gravitational centers of our ellipse.

### The Hyperbola's Identity

Now, we turn to the hyperbola. We are told its transverse axis length is $\sqrt{2}$.

The length of the transverse axis is $2a_h$, so $2a_h = \sqrt{2}$, which gives us $a_h = \frac{1}{\sqrt{2}}$. Because the hyperbola shares the same foci as the ellipse, its foci must also be at $(\pm 1, 0)$.

For a hyperbola, the foci are at $(\pm a_h e_h, 0)$. Thus, $a_h e_h = 1$. Substituting our known $a_h$, we get:

$$\frac{1}{\sqrt{2}} e_h = 1 \Rightarrow e_h = \sqrt{2}$$

Now, we calculate the hyperbola's conjugate axis parameter $b_h^2$. The formula is $b_h^2 = a_h^2(e_h^2 - 1)$.

Substituting our values, we get:

$$b_h^2 = \left(\frac{1}{\sqrt{2}}ight)^2 ((\sqrt{2})^2 - 1) = \frac{1}{2}(2 - 1) = \frac{1}{2}$$

### The Final Construction

With $a_h^2 = \frac{1}{2}$ and $b_h^2 = \frac{1}{2}$, the equation of our hyperbola is:

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

This simplifies beautifully to **$2x^2 - 2y^2 = 1$**. This is the equation that governs every point on our hyperbola.

To find which point does not lie on it, we test the options. Let us check the point $(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}})$.

Substituting these into our equation:

$$2\left(\sqrt{\frac{3}{2}}ight)^2 - 2\left(\frac{1}{\sqrt{2}}ight)^2 = 2\left(\frac{3}{2}ight) - 2\left(\frac{1}{2}ight) = 3 - 1 = 2$$

Since $2 
eq 1$, this point clearly does not satisfy the equation. We have successfully navigated the geometry and the algebra to find our answer.

Animated Solution for Mathematics - Conic Sections: A hyperbola having the transverse axis of length $2$ has the same foci as that of the ellipse of $3 x^{2} + 4 y^{2} = 12$ , then this hyperbola does not pass through which of the following points?

Select Answer:

Visualized Solution

Analyze the Ellipse Equation

Ellipse Eccentricity Formula

Calculate Ellipse Eccentricity

Locate the Shared Foci

Hyperbola Transverse Axis

Hyperbola Foci Condition

Find Hyperbola Eccentricity

Calculate $b_{h}^{2}$

Construct the Hyperbola Equation

Testing the Points

Final Summary

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

Analyzing the Setup

Unmasking the Ellipse

The Hyperbola's Identity

The Final Construction

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