If the length of the latus rectum of the ellipse $x^2 + 4y^2 + 2x + 8y - \lambda = 0$ is 4, and $l$ is the length of its major axis, then $\lambda + l$ is equal to

### Analyzing the Setup To uncover the geometry of the ellipse, we begin with the given equation: $$x^2 + 4y^2 + 2x + 8y - \lambda = 0$$ We group the $x$ terms and the $y$ terms to prepare for completing the square: $$(x^2 + 2x) + (4y^2 + 8y) = \lambda$$ ### Completing the Square For the $x$ terms, we add $1$ to form a perfect square. For the $y$ terms, we factor out $4$ to get $4(y^2 + 2y)$ and add $1$ inside the bracket: $$(x^2 + 2x + 1) + 4(y^2 + 2y + 1) = \lambda + 1 + 4(1)$$ This simplifies to the following expression: $$(x+1)^2 + 4(y+1)^2 = \lambda + 5$$ ### Standard Form Conversion Dividing both sides by $\lambda + 5$, we obtain the standard form of the ellipse: $$\frac{(x+1)^2}{\lambda + 5} + \frac{(y+1)^2}{\frac{\lambda + 5}{4}} = 1$$ From this, we identify the parameters $a^2 = \lambda + 5$ and $b^2 = \frac{\lambda + 5}{4}$. Consequently, $a = \sqrt{\lambda + 5}$ and $b = \frac{\sqrt{\lambda + 5}}{2}$. ### Applying the Latus Rectum Constraint The length of the latus rectum is given as $4$. Using the formula $\frac{2b^2}{a}$, we substitute our values: $$\frac{2 \left( \frac{\lambda + 5}{4} \right)}{\sqrt{\lambda + 5}} = 4$$ Simplifying the expression leads to: $$\frac{\sqrt{\lambda + 5}}{2} = 4$$ Solving for $\lambda$, we find $\sqrt{\lambda + 5} = 8$, which implies $\lambda + 5 = 64$, or **$\lambda = 59$**. ### Final Calculation The length of the major axis $l$ is defined as $2a$: $$l = 2\sqrt{\lambda + 5} = 2\sqrt{64} = 16$$ Adding the values together, we reach the final result: $$\lambda + l = 59 + 16 = \mathbf{75}$$

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Animated Solution for Mathematics - Conic Sections: If the length of the latus rectum of the ellipse $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$ is 4, and $l$ is the length of its major axis, then $λ + l$ is equal to

Q: How do I know which denominator is $a^2$ and which is $b^2$?

In the standard form of an ellipse, $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, the convention is that $a$ is the semi-major axis and $b$ is the semi-minor axis. Therefore, $a^2$ is always the larger denominator. If you find that the denominator under the $x$-term is larger, your ellipse is horizontal; if the denominator under the $y$-term is larger, it is vertical.

Enter Numerical Value:

Visualized Solution

Given Equation

Given equation: $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$
Goal: Convert to standard form $\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1$

Grouping Terms

Group terms: $(x^{2} + 2 x) + (4 y^{2} + 8 y) = λ$

Completing Square for $x$

Focus on x-terms: $x^{2} + 2 x$
Add $1$ to complete the square: $(x^{2} + 2 x + 1)$

Completing Square for $y$

Focus on y-terms: $4 (y^{2} + 2 y)$
Add $1$ inside bracket: $4 (y^{2} + 2 y + 1)$
This adds $4 \times 1 = 4$ to the left side.

The Balanced Equation

Add constants to the right side: $(x + 1)^{2} + 4 (y + 1)^{2} = λ + 1 + 4$
Simplified balanced equation: $(x + 1)^{2} + 4 (y + 1)^{2} = λ + 5$

Standard Form

Divide by $(λ + 5)$ to make RHS $1$
$\frac{( x + 1 ) ^{2}}{λ + 5} + \frac{4 ( y + 1 ) ^{2}}{λ + 5} = 1$
Standard form: $\frac{( x + 1 ) ^{2}}{λ + 5} + \frac{( y + 1 ) ^{2}}{\frac{λ + 5}{4}} = 1$

Identifying $a^{2}$ and $b^{2}$

Compare with $\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1$
$a^{2} = λ + 5 ⟹ a = λ + 5$
$b^{2} = \frac{λ + 5}{4} ⟹ b = \frac{λ + 5}{2}$

Latus Rectum Formula

Length of Latus Rectum (L.R.) $= \frac{2 b ^{2}}{a}$
Given: L.R. $= 4$

Substituting Values

Substitute $a$ and $b^{2}$ :
$\frac{2 ( \frac{λ + 5}{4} )}{λ + 5} = 4$

Simplifying the Expression

Simplify numerator: $\frac{λ + 5}{2}$
$\frac{\frac{λ + 5}{2}}{λ + 5} = 4$
$\frac{λ + 5}{2} = 4$

Solving for $λ$

Multiply by $2$ : $λ + 5 = 8$
Square both sides: $λ + 5 = 64$
$λ = 59$

Length of Major Axis $l$

Major axis length $l = 2 a$
We know $a = λ + 5$
$a = 59 + 5 = 64 = 8$

Calculating $l$

$l = 2 \times 8 = 16$

Final Calculation $λ + l$

Calculate $λ + l$ :
$λ + l = 59 + 16$
$λ + l = 75$

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

Solution Diagram

Analyzing the Setup

To uncover the geometry of the ellipse, we begin with the given equation:

x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0

We group the

x

terms and the

y

terms to prepare for completing the square:

(x^{2} + 2 x) + (4 y^{2} + 8 y) = λ

Completing the Square

For the

x

terms, we add

1

to form a perfect square. For the

y

terms, we factor out

4

to get

4 (y^{2} + 2 y)

and add

1

inside the bracket:

(x^{2} + 2 x + 1) + 4 (y^{2} + 2 y + 1) = λ + 1 + 4 (1)

This simplifies to the following expression:

(x + 1)^{2} + 4 (y + 1)^{2} = λ + 5

Standard Form Conversion

Dividing both sides by

λ + 5

, we obtain the standard form of the ellipse:

\frac{( x + 1 ) ^{2}}{λ + 5} + \frac{( y + 1 ) ^{2}}{\frac{λ + 5}{4}} = 1

From this, we identify the parameters

a^{2} = λ + 5

and

b^{2} = \frac{λ + 5}{4}

. Consequently,

a = λ + 5

and

b = \frac{λ + 5}{2}

Applying the Latus Rectum Constraint

The length of the latus rectum is given as

4

. Using the formula

\frac{2 b ^{2}}{a}

, we substitute our values:

\frac{2 ( \frac{λ + 5}{4} )}{λ + 5} = 4

Simplifying the expression leads to:

\frac{λ + 5}{2} = 4

Solving for

λ

, we find

λ + 5 = 8

, which implies

λ + 5 = 64

, or $λ = 59$ .

Final Calculation

The length of the major axis

l

is defined as

2 a

l = 2 λ + 5 = 264 = 16

Adding the values together, we reach the final result:

λ + l = 59 + 16 = 75

Common Traps & FAQs

Wait, why do we add constants to the right side when completing the square? Isn't that just changing the equation?

That is a fantastic question! Think of the equation as a balanced scale. When we 'complete the square' on the left side, we are essentially adding a specific value to make a trinomial a perfect square. If we add that value to the left, the scale tips. To keep the equation true, we must add the exact same value to the right side. It is not changing the equation; it is simply rewriting it in a more useful, revealing form.

How do I know which denominator is

a^{2}

and which is

b^{2}

In the standard form of an ellipse,

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1

, the convention is that

a

is the semi-major axis and

b

is the semi-minor axis. Therefore,

a^{2}

is always the larger denominator. If you find that the denominator under the

x

-term is larger, your ellipse is horizontal; if the denominator under the

y

-term is larger, it is vertical.

Is the formula for the length of the latus rectum always

\frac{2 b ^{2}}{a}

Yes, it is! The latus rectum is the chord passing through the focus perpendicular to the major axis. Regardless of whether the ellipse is horizontal or vertical, the length is always

2 \times \frac{( semi-minor axis ) ^{2}}{semi-major axis}

. Just ensure you correctly identify which variable is the semi-major axis (

a

) and which is the semi-minor axis (

b

Animated Solution for Mathematics - Conic Sections: If the length of the latus rectum of the ellipse $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$ is 4, and $l$ is the length of its major axis, then $λ + l$ is equal to

Enter Numerical Value:

Visualized Solution

Given Equation

Grouping Terms

Completing Square for $x$

Completing Square for $y$

The Balanced Equation

Standard Form

Identifying $a^{2}$ and $b^{2}$

Latus Rectum Formula

Substituting Values

Simplifying the Expression

Solving for $λ$

Length of Major Axis $l$

Calculating $l$

Final Calculation $λ + l$

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

Analyzing the Setup

Completing the Square

Standard Form Conversion

Applying the Latus Rectum Constraint

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 :

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $(0, 53)$ , then the length of its latus rectum 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

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 :

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is $\frac{3}{2}$ units, then its eccentricity is :

The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, - 3)$ and eccentricity is $\frac{4}{5}$ , is

For some $θ \in (0, \frac{π}{2})$ , if the eccentricity of the hyperbola, $x^{2} - y^{2} sec^{2} θ = 10$ is $5$ times the eccentricity of the ellipse, $x^{2} sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is :

Let the ellipse $E : \frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1$ and the hyperbola $H : \frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$ , then the value of $24 (e + L)$ is :

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?

Enter Numerical Value:

Visualized Solution

Given Equation

Grouping Terms

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

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

The Balanced Equation

Standard Form

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Latus Rectum Formula

Substituting Values

Simplifying the Expression

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

Length of Major Axis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

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

Final Calculation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ+l

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

Analyzing the Setup

Completing the Square

Standard Form Conversion

Applying the Latus Rectum Constraint

Final Calculation

Similar Questions

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

.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; } The length of the latus-rectum of the ellipse, whose foci are (2,5) and (2,−3) and eccentricity is 54​, is

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?

Completing Square for $x$

Completing Square for $y$

Identifying $a^{2}$ and $b^{2}$

Solving for $λ$

Length of Major Axis $l$

Calculating $l$

Final Calculation $λ + l$

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 :

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