Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the

Q: Wait, why is the distance between the foci $2ae$ and not just $ae$?

Great question! The foci are located at $(ae, 0)$ and $(-ae, 0)$. The distance between them is the difference in their x-coordinates: $ae - (-ae) = 2ae$. Always remember to account for both sides of the origin!

Q: Why can't we just use the eccentricity formula $e = \sqrt{1 - \frac{b^2}{a^2}}$ directly?

You absolutely can! However, working with square roots often leads to algebraic errors. Using the relation $ae = b$ allows you to work with squares ($a^2e^2 = b^2$) directly, which is much cleaner and less prone to mistakes.

Q: How do I know which point to check first in the options?

Once you have the equation $\frac{x^2}{64} + \frac{y^2}{32} = 1$, you simply plug the coordinates $(x, y)$ from each option into the equation. The point that results in a sum of $1$ is your winner. It's a straightforward verification step!

Question

Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

Options:

(A) $(4\sqrt{2}, 2\sqrt{3})$

(B) $(4\sqrt{3}, 2\sqrt{2})$

(C) $(4\sqrt{3}, 2\sqrt{3})$

(D) $(4\sqrt{2}, 2\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the world of coordinate geometry! Today, we are going to peel back the layers of an ellipse.

We consider an ellipse centered at the origin with its major axis along the $x$-axis. We define its standard form as:

$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad 	ext{where } a > b
$$

### Decoding the Clues

The problem provides two vital pieces of information, the 'DNA' of our ellipse. First, the length of the latus rectum is $8$.

Using the standard formula for the latus rectum, we have:

$$
\frac{2b^2}{a} = 8 \Rightarrow b^2 = 4a
$$

Next, we are told the distance between the foci is equal to the length of the minor axis. The distance between the foci is $2ae$, and the length of the minor axis is $2b$.

Equating these, we get $2ae = 2b$, which simplifies to:

$$
ae = b
$$

### The Algebraic Dance

Now, we utilize the fundamental eccentricity relation for an ellipse:

$$
b^2 = a^2(1 - e^2)
$$

Expanding this, we obtain $b^2 = a^2 - a^2e^2$. Since we know $ae = b$, it follows that $a^2e^2 = b^2$.

Substituting this into our relation, the equation transforms into:

$$
b^2 = a^2 - b^2 \Rightarrow a^2 = 2b^2
$$

We now have a system of two equations: $b^2 = 4a$ and $a^2 = 2b^2$. Substituting the first into the second:

$$
a^2 = 2(4a) \Rightarrow a^2 = 8a
$$

Since $a 
eq 0$, we divide by $a$ to find **$a = 8$**. Consequently, we find **$b^2 = 4(8) = 32$**.

### Final Calculation

Our ellipse is now fully defined by the equation:

$$
\frac{x^2}{64} + \frac{y^2}{32} = 1
$$

To verify a point such as $(4\sqrt{3}, 2\sqrt{2})$, we substitute the coordinates into our equation:

$$
\frac{(4\sqrt{3})^2}{64} + \frac{(2\sqrt{2})^2}{32} = \frac{48}{64} + \frac{8}{32}
$$

Simplifying the fractions, we get:

$$
\frac{3}{4} + \frac{1}{4} = 1
$$

The point satisfies the equation perfectly. You have successfully navigated the geometry and the algebra to define the ellipse.

Select Answer:

Visualized Solution

Equation of the Ellipse

Latus Rectum Formula

Relation between $a$ and $b$

Foci and Minor Axis

Simplifying Foci Condition

Eccentricity Relation

Substituting $a e = b$

Solving for $a^{2}$

Finding the value of $a$

Equation of the Ellipse

Checking the Points

Evaluating the Point

Final Answer

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

Analyzing the Setup

Decoding the Clues

The Algebraic Dance

Final Calculation

Similar Questions

Let $S$ and $S^{'}$ be the foci of the ellipse and $B$ be any one of the extremities of its minor axis. If $S^{'} B S$ is a right angled triangle with right angle at $B$ and area $(S^{'} B S) = 8$ sq. units, then the length of a latus rectum of the ellipse is :

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Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1$ . If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____.

Select Answer:

Visualized Solution

Equation of the Ellipse

Latus Rectum Formula

Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Foci and Minor Axis

Simplifying Foci Condition

Eccentricity Relation

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ae=b

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Finding the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Equation of the Ellipse

Checking the Points

Evaluating the Point

Final Answer

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

Analyzing the Setup

Decoding the Clues

The Algebraic Dance

Final Calculation

Similar Questions

An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through which of the following points ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An ellipse has its center at (1,−2), one focus at (3,−2) and one vertex at (5,−2). Then the length of its latus rectum is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the foci of the ellipse 16x2​+7y2​=1 and the hyperbola 144x2​−αy2​=251​ coincide. Then the length of the latus rectum of the hyperbola is:-

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

Relation between $a$ and $b$

Substituting $a e = b$

Solving for $a^{2}$

Finding the value of $a$

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :

Let the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-