A latus rectum of an ellipse is a chord passing through the foci and perpendicular to the major axis. Suppose that an ellipse has major and minor axes of lengths 2a and 2b, where a>b.
a) (3 marks) Show that the length of a latus rectum of an ellipse is 2b2/a. Hint: The foci of an ellipse lie on the major axis,units from the centre of the ellipse, witha22 having equation Give all answers in exact form
(b) (3 marks) Find the endpoints of the latera recta (plural of latus rectum) for the ellipse 13r2 + 26x +4y2-12y -30.
The correct answer and explanation is :
Alright, let’s break it down carefully:
Part (a): Show that the length of the latus rectum is ( \frac{2b^2}{a} )
The standard form of an ellipse centered at the origin with a horizontal major axis is:
[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\quad \text{where} \quad a > b > 0
]
The foci are located at ( (c, 0) ) and ( (-c, 0) ), where:
[
c = \sqrt{a^2 – b^2}
]
The latus rectum is a vertical line segment passing through a focus and perpendicular to the major axis.
Suppose we focus on the focus at ( (c, 0) ).
Since the latus rectum is vertical at ( x = c ), substitute ( x = c ) into the ellipse equation:
[
\frac{c^2}{a^2} + \frac{y^2}{b^2} = 1
]
We know ( c^2 = a^2 – b^2 ), so:
[
\frac{a^2 – b^2}{a^2} + \frac{y^2}{b^2} = 1
]
[
1 – \frac{b^2}{a^2} + \frac{y^2}{b^2} = 1
]
[
\frac{y^2}{b^2} = \frac{b^2}{a^2}
]
[
y^2 = \frac{b^4}{a^2}
]
[
y = \pm \frac{b^2}{a}
]
Thus, the endpoints of the latus rectum at focus ( (c, 0) ) are:
[
\left(c, \frac{b^2}{a}\right) \quad \text{and} \quad \left(c, -\frac{b^2}{a}\right)
]
The length of the latus rectum is the vertical distance between these two points:
[
\text{Length} = \left( \frac{b^2}{a} – \left(-\frac{b^2}{a}\right) \right) = \frac{2b^2}{a}
]
✅ Hence, proved that the length of a latus rectum is ( \boxed{\frac{2b^2}{a}} ).
Part (b): Find the endpoints of the latera recta for the ellipse ( 13x^2 + 26x + 4y^2 – 12y – 30 = 0 )
First, rewrite the given ellipse in standard form:
Group ( x )-terms and ( y )-terms:
[
13(x^2 + 2x) + 4(y^2 – 3y) = 30
]
Complete the square:
- For ( x )-terms: ( x^2 + 2x \to (x+1)^2 – 1 )
- For ( y )-terms: ( y^2 – 3y \to (y – \frac{3}{2})^2 – \frac{9}{4} )
Substituting:
[
13\left((x+1)^2 – 1\right) + 4\left(\left(y – \frac{3}{2}\right)^2 – \frac{9}{4}\right) = 30
]
Expand:
[
13(x+1)^2 – 13 + 4\left(y-\frac{3}{2}\right)^2 – 9 = 30
]
[
13(x+1)^2 + 4\left(y-\frac{3}{2}\right)^2 – 22 = 30
]
[
13(x+1)^2 + 4\left(y-\frac{3}{2}\right)^2 = 52
]
Divide by 52:
[
\frac{(x+1)^2}{4} + \frac{(y – \frac{3}{2})^2}{13} = 1
]
Now it’s in standard form:
[
\frac{(x+1)^2}{4} + \frac{(y-\frac{3}{2})^2}{13} = 1
]
Compare with:
[
\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1
]
Thus:
- Center ( (h, k) = (-1, \frac{3}{2}) )
- ( a^2 = 4, \quad b^2 = 13 )
Notice b > a, so the major axis is vertical!
Step 1: Find ( c ) (distance to foci):
[
c = \sqrt{b^2 – a^2} = \sqrt{13 – 4} = \sqrt{9} = 3
]
Thus, the foci are at:
[
(h, k \pm c) = \left(-1, \frac{3}{2} \pm 3\right)
]
So:
- Upper focus: ( \left(-1, \frac{9}{2}\right) )
- Lower focus: ( \left(-1, \frac{-3}{2}\right) )
Step 2: Find endpoints of the latera recta:
Since the major axis is vertical, the latus rectum at each focus is horizontal, passing through a focus.
Substitute ( y = \frac{9}{2} ) into the ellipse equation to find corresponding ( x )-values:
Start with:
[
\frac{(x+1)^2}{4} + \frac{(y-\frac{3}{2})^2}{13} = 1
]
At ( y = \frac{9}{2} ):
[
\frac{(x+1)^2}{4} + \frac{\left(\frac{9}{2} – \frac{3}{2}\right)^2}{13} = 1
]
Simplify:
[
\frac{(x+1)^2}{4} + \frac{(3)^2}{13} = 1
]
[
\frac{(x+1)^2}{4} + \frac{9}{13} = 1
]
[
\frac{(x+1)^2}{4} = \frac{4}{13}
]
[
(x+1)^2 = \frac{16}{13}
]
[
x+1 = \pm \frac{4}{\sqrt{13}}
]
Thus:
[
x = -1 \pm \frac{4}{\sqrt{13}}
]
Similarly, at ( y = -\frac{3}{2} ) (lower focus):
Substitute ( y = -\frac{3}{2} ) into the ellipse:
[
\frac{(x+1)^2}{4} + \frac{\left(-\frac{3}{2} – \frac{3}{2}\right)^2}{13} = 1
]
Simplify:
[
\frac{(x+1)^2}{4} + \frac{(-3)^2}{13} = 1
]
[
\frac{(x+1)^2}{4} + \frac{9}{13} = 1
]
[
\frac{(x+1)^2}{4} = \frac{4}{13}
]
Thus the same computation for ( x ):
[
x = -1 \pm \frac{4}{\sqrt{13}}
]
Final Answer:
- Endpoints of upper latus rectum:
[
\left( -1 + \frac{4}{\sqrt{13}}, \frac{9}{2} \right) \quad \text{and} \quad \left( -1 – \frac{4}{\sqrt{13}}, \frac{9}{2} \right)
] - Endpoints of lower latus rectum:
[
\left( -1 + \frac{4}{\sqrt{13}}, -\frac{3}{2} \right) \quad \text{and} \quad \left( -1 – \frac{4}{\sqrt{13}}, -\frac{3}{2} \right)
]
Short Explanation (About 300 words)
The latus rectum of an ellipse is a chord that passes through a focus and is perpendicular to the major axis. In this problem, we were given a conic section in a general quadratic form. We first rewrote it into standard form by completing the square. After completing the square, we obtained the standard ellipse equation, recognizing that the major axis is vertical because ( b^2 > a^2 ).
From the standard form, we identified the center ( (-1, 3/2) ) and semi-axes lengths ( a=2 ) and ( b=\sqrt{13} ). To find the foci, we computed the focal distance ( c = 3 ), and so the foci are located vertically above and below the center.
To find the endpoints of the latus recta, we substituted the ( y )-coordinates of the foci into the ellipse equation and solved for corresponding ( x )-values. Since the latus rectum is horizontal in this case (vertical major axis), the endpoints are horizontally displaced from the focus. Specifically, at each focus, the two endpoints are ( \frac{8}{\sqrt{13}} ) units apart horizontally.
Thus, the endpoints were expressed exactly in terms of square roots: ( -1 \pm \frac{4}{\sqrt{13}} ) for ( x ), with corresponding ( y )-values ( \frac{9}{2} ) and ( -\frac{3}{2} ). This detailed and exact method ensures full understanding of the geometric properties of ellipses.