Q.A beam is supported at its ends by supports which are 12 metres apart. Since the load is concentrated at the centre of the beam there is a deflection of 3 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm?
Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
95% · 38/40 Questions
✓ Free question
Concept understanding — Length of the Latus Rectum
What is the Latus Rectum? A First Look
Imagine you draw a parabola, ellipse, or hyperbola — a conic section. Each of these curves has a special line segment that cuts straight across it, passing through the focus and running parallel to the directrix. That segment is the latus rectum.
The name comes from Latin: latus means "side" and rectum means "straight." So it's literally the "straight side" of the curve. For a student meeting it for the first time, think of it as the width of the curve at the focus — how wide the opening is right at that special point.
Intuition First
Take a parabola that opens upward, like y2=4ax. Its focus is at (a,0). If you draw a vertical line through that focus, it will hit the parabola at two points — one above, one below. The distance between those two intersection points is the length of the latus rectum.
Why does this matter? Because it tells you how "fat" or "skinny" the curve is. A larger latus rectum means a wider opening; a smaller one means a tighter, narrower curve. For ellipses and hyperbolas, it also relates directly to the shape's eccentricity.
The Precise Statement
For a conic section with focus at (a,0) and directrix x=−a (standard parabola y2=4ax):
Length of latus rectum=4a
But that's just the parabola. Here are the exact formulas for all three conics:
Conic
Standard Equation
Length of Latus Rectum
Parabola
y2=4ax
4a
Ellipse
a2x2+b2y2=1 (a > b)
a2b2
Hyperbola
a2x2−b2y2=1
a2b2
Notice something interesting: the ellipse and hyperbola share the same formula. That's because both have two foci and the latus rectum is defined through either focus.
Why That Formula? A Quick Derivation
For the parabola y2=4ax, the focus is at (a,0). A line through the focus parallel to the directrix is vertical (since the directrix is vertical x=−a). So the line is x=a.
Substitute x=a into y2=4ax:
y2=4a(a)=4a2
y=±2a
The two intersection points are (a,2a) and (a,−2a). The distance between them is 4a. That's it.
Tip
For any conic, the latus rectum is always found by substituting the focus's x-coordinate into the equation and solving for y. The distance between the two y-values is the length.
Common Mistake to Avoid
Watch out
Do not confuse the latus rectum with the focal width — they are the same thing. But some students mistakenly think the latus rectum is the distance from the focus to the curve. It is not. It is the full chord through the focus, perpendicular to the axis.
Also, for ellipses and hyperbolas, there are two latera recta — one through each focus. They have the same length.
Why It Matters in Exams
You will be asked to:
Find the length of the latus rectum given the equation of a conic.
Use it to find a or b when the length is given.
Relate it to eccentricity (especially for ellipses: e=1−a2b2, so the latus rectum a2b2 ties directly to e).
Important
For a parabola, the latus rectum is 4a. For an ellipse or hyperbola, it is a2b2. Memorize these — they appear in nearly every conic section problem.
One Final Intuition
Picture a parabola like a U-shaped slide. The latus rectum is the width of that slide exactly at the level of the focus. For an ellipse, imagine an oval — the latus rectum is the length of the vertical line segment that passes through one focus and touches the ellipse at two points. It's a concrete, measurable property that captures how "stretched" or "squashed" the curve is at its focal point.
Modelling the deflected beam as an upward-opening parabola with its vertex at the point of maximum deflection, the support points (where deflection is zero) fix the parabola's equation, which can then be solved for the position where the deflection equals 1 cm.
✓Final answer
The deflection is 1 cm at a distance 26≈4.90 m from the centre.
Model the deflected beam as a parabola with vertex at the point of maximum (3 cm) deflection; use the support points (zero deflection) to fix the parabola, then solve for the position where the deflection is 1 cm.
Parabola with vertex at origin, axis vertical, opening upward: x2=4ay.
Place the origin at the centre of the beam, at the point of maximum deflection (3 cm). Let y measure how much the beam has risen back toward the support level as we move away from the centre (so y=0 at the centre, and y=3 cm at each support, where deflection is 0).
The supports are 12 m apart, so each support is at horizontal distance x=6 m from the centre. At the support, deflection is 0, i.e. the beam has risen the full 3 cm, so y=3 there.
Substitute the point (x,y)=(6,3) (working in metres for x and cm for y, consistent with the given data) into x2=4ay:
62=4a(3)⇒36=12a⇒a=3
Parabola: x2=12y.
We need the point where the deflection is 1 cm, i.e. the beam has risen 3−1=2 cm from the centre, so y=2:
x2=12(2)=24⇒x=24=26≈4.899 m
Self-check: at x=26, y=12(26)2=1224=2 cm risen ⇒ deflection =3−2=1 cm ✓, matching the requirement.
✓Final answer
The deflection is 1 cm at a horizontal distance x=26≈4.90 m from the centre of the beam.