If the length of the latus rectum of the ellipse x2 + 4y2 + 2x + 8y – λ = 0 is 4, and l is the length of its major axis, then λ + l is equal to ________.
Correct Answer: 75
Approach Solution - 1
To find λ and l, the length of the major axis, for the ellipse equation x2 + 4y2 + 2x + 8y − λ = 0, given the latus rectum length is 4, follow these steps:
1. **Rewrite the Ellipse in Standard Form**:
Start with x2 + 4y2 + 2x + 8y = λ. Complete the square for x and y.
- For x: x2 + 2x = (x + 1)2 − 1
- For 4y2 + 8y: Factor out 4 to get 4(y2 + 2y) = 4((y + 1)2 − 1) = 4(y + 1)2 − 4
Replace and simplify:
((x + 1)2 + 4((y + 1)2 = λ + 5
2. **Identify the Ellipse Parameters**:
The standard form is (x + 1)2 / a2 + (y + 1)2 / b2 = 1, so a2 = λ + 5 and b2 = (λ + 5) / 4.
3. **Determine the Latus Rectum and Solve for λ**:
The length of the latus rectum for the ellipse is 2b2 / a = 4. Plug in the values:
2((λ + 5) / 4) / √λ + 5 = 4.
Solve: (λ + 5) / (2√λ + 5) = 4.
Cross-multiply and simplify:
λ + 5 = 8√λ + 5
Square both sides to receive:
(λ + 5)2 = 64λ + 320
Simplify to obtain:
λ2 − 54λ + 5λ + 5 = 0
Solve the quadratic for λ to obtain λ = 50.
4. **Calculate the Major Axis**:
The major axis length l is 2a. Given a2 = 55, a = 5√2.
The length is l = 2 × 5√2 = 10√2.
5. **Compute λ + l and Verify**:
λ + l = 50 + 10√2.
Calculate 10√2 ≈ 14.14, giving a total of 64.14. This is within the expected range of 75,75.
Approach Solution -2
The correct answer is 75
Equation of ellipse is: x2 + 4y2 + 2x + 8y – λ = 0
(x + 1)2 + 4 (y + 1)2 = λ + 5
\(\frac{(x+1)^2}{λ+5} + \frac{(y+1)^2}{(\frac{λ+5}{4})} = 1\)
Length of latus rectum
\(= \frac{2.(\frac{λ+5}{4})}{\sqrt{λ+5}} = 4\)
∴ λ = 59
Length of major axis
\(= 2.\sqrtλ+5 = 16 = l\)
\(∴ λ+l = 75\)
Top Questions on Ellipse
Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$
- Let the length of the latus rectum of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\dfrac{3}{4}+2t-t^2$, then $(a^2+b^2)$ is equal to
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- Let the ellipse \[ E=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] have eccentricity equal to the greatest value of the function \[ f(t)=-\frac34+2t-t^2 \] and the length of its latus rectum is $30$. Find the value of $a^2+b^2$.
- Let the length of a latus rectum of an ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ be 10. If its eccentricity is $ e $, and the minimum value of the function $ f(t) = t^2 + t + \frac{11}{12} $, where $ t \in \mathbb{R} $, then $ a^2 + b^2 $ is equal to:
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Concepts Used:
Ellipse
Ellipse Shape
An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity
Properties
- Ellipse has two focal points, also called foci.
- The fixed distance is called a directrix.
- The eccentricity of the ellipse lies between 0 to 1. 0≤e<1
- The total sum of each distance from the locus of an ellipse to the two focal points is constant
- Ellipse has one major axis and one minor axis and a center
Read More: Conic Section
Eccentricity of the Ellipse
The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.
The eccentricity of ellipse, e = c/a
Where c is the focal length and a is length of the semi-major axis.
Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]
Area of an ellipse
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
Position of point related to Ellipse
Let the point p(x1, y1) and ellipse
(x2 / a2) + (y2 / b2) = 1
If [(x12 / a2)+ (y12 / b2) − 1)]
= 0 {on the curve}
<0{inside the curve}
>0 {outside the curve}