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Parabola – Definition, Standard Equations and Properties

🎓 Class 11 Mathematics CBSE Theory Ch 10 — Conic Sections ⏱ ~30 min
🌐 Language: [gtranslate]

This MCQ module is based on: Parabola – Definition, Standard Equations and Properties

This mathematics assessment will be based on: Parabola – Definition, Standard Equations and Properties
Targeting Class 11 level in Coordinate Geometry, with Advanced difficulty.

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10.4 Parabola

Definition
A parabolai is the locus of all points in the plane that are equidistant from a fixed point \(F\) (called the focus) and a fixed line \(l\) (called the directrix), with \(F\) not on \(l\).

10.4.1 Standard equation — \(y^2=4ax\)

Let the focus be \(F(a,0)\) and directrix \(x=-a\) with \(a>0\). For any point \(P(x,y)\) on the curve, the focus distance equals the perpendicular distance to the directrix:

\(\sqrt{(x-a)^2+y^2}=|x+a|.\) Squaring and simplifying, \((x-a)^2+y^2=(x+a)^2\Rightarrow y^2=4ax.\)

Standard form
The parabola \(y^2=4ax\) has vertex \(V(0,0)\), focus \(F(a,0)\), directrix \(x=-a\) and axis along the \(x\)-axis. It opens to the right.
xy F(a,0) x = −a (directrix) V Latus rectum axis
Fig 10.4 — Standard parabola \(y^2=4ax\) with vertex \(V\), focus \(F(a,0)\), directrix \(x=-a\).

10.4.2 The four standard orientations

Depending on which way the parabola opens, we have four standard forms (vertex at origin, axis along a coordinate axis):

EquationOpensFocusDirectrixAxis
\(y^2=4ax\)right (\(a>0\))\((a,0)\)\(x=-a\)\(x\)-axis
\(y^2=-4ax\)left\((-a,0)\)\(x=a\)\(x\)-axis
\(x^2=4ay\)up\((0,a)\)\(y=-a\)\(y\)-axis
\(x^2=-4ay\)down\((0,-a)\)\(y=a\)\(y\)-axis
y² = 4ax y² = −4ax x² = 4ay x² = −4ay
Fig 10.5 — The four standard parabolas.

10.4.3 Latus rectum

The latus rectumi is the chord through the focus perpendicular to the axis. For \(y^2=4ax\), at \(x=a\): \(y^2=4a^2\Rightarrow y=\pm 2a\). Hence the latus rectum is the segment from \((a,2a)\) to \((a,-2a)\) and has length \(4a\). Its endpoints are \(L(a,2a)\) and \(L'(a,-2a)\).

Example 5
Find the focus, axis, directrix and length of the latus rectum of the parabola \(y^2=12x\).
Compare \(y^2=4ax\): \(4a=12\Rightarrow a=3\). Focus \((3,0)\), axis \(x\)-axis, directrix \(x=-3\), latus rectum \(=12\).
Example 6
Find the equation of the parabola with vertex at origin, axis along \(y\)-axis, passing through \((5,2)\) and opening upward.
Form: \(x^2=4ay\). Through \((5,2)\): \(25=8a\Rightarrow a=\tfrac{25}{8}\). Equation: \(x^2=\tfrac{25}{2}y\) or \(2x^2=25y\).
Example 7
Find the equation of the parabola with focus \((0,-3)\) and directrix \(y=3\).
Vertex at midpoint of perpendicular from focus to directrix \(=(0,0)\). \(a=3\), opens down. Equation: \(x^2=-12y.\)
Example 8 — Application
A beam of parallel rays from infinity hits a parabolic reflector with equation \(y^2=20x\). Where is the focus (the point where rays converge)?
\(4a=20\Rightarrow a=5\). Focus is \((5,0)\).
Example 9 — Focal distance
For a point \(P(x_0,y_0)\) on \(y^2=4ax\), show that the distance from \(P\) to the focus equals \(x_0+a\).
By the definition (focus–directrix), focal distance = perpendicular distance from \(P\) to directrix \(x=-a\), which is \(x_0+a.\)
Activity 10.2 — The string-and-pin parabola
Predict: You fix a pin at a point \(F\), draw a horizontal line \(l\) some distance below, and trace a curve such that every traced point is equidistant from \(F\) and \(l\). What shape will appear?
  1. On graph paper, mark focus \(F(0,2)\) and directrix \(y=-2\).
  2. Using a set-square or a string looped through \(F\), plot 10 points equidistant from \(F\) and the line.
  3. Join the points smoothly — does it match the equation \(x^2=4(2)y=8y\)?
  4. Measure the focal chord through \(F\) perpendicular to the axis — verify it has length \(4a=8\).
Insight: The traced locus is a parabola opening upward. The vertex sits midway between focus and directrix, confirming \(a=2\). The perpendicular focal chord (latus rectum) has length \(4a=8\) — the shortest focal chord.

In-text Exercises on the Parabola

Q1. Find the coordinates of focus, axis, directrix and length of the latus rectum of the parabola \(x^2=6y\).
\(4a=6\Rightarrow a=\tfrac{3}{2}\). Focus \((0,\tfrac{3}{2})\); axis: \(y\)-axis; directrix \(y=-\tfrac{3}{2}\); LR \(=6\).
Q2. Find the equation of the parabola with focus \((6,0)\) and directrix \(x=-6\).
Vertex at origin, \(a=6\), opens right: \(y^2=24x.\)
Q3. Find the equation of the parabola whose axis is the \(y\)-axis, vertex is origin and which passes through \((3,-4)\).
\(x^2=-4ay\) (opens down because \(y\) is negative). At \((3,-4)\): \(9=16a\Rightarrow a=\tfrac{9}{16}\). Equation: \(x^2=-\tfrac{9}{4}y\) i.e. \(4x^2+9y=0.\)
Q4. Find the equation of the parabola with focus \((-2,0)\) and directrix \(x=2\).
Opens left, \(a=2\): \(y^2=-8x.\)
Competency-Based Questions — Parabolic Reflectors
An engineering team designs a parabolic dish antenna. A cross-section of the dish (opening upward) is modelled in a coordinate plane with the dish's vertex at the origin and axis along the \(y\)-axis. The dish is 8 m wide at the rim and 2 m deep at the centre.
Q1. Find the equation of the parabolic cross-section.
L3 Apply
Form: \(x^2=4ay\). Rim at \((\pm 4, 2)\): \(16=8a\Rightarrow a=2\). Equation: \(x^2=8y.\)
Q2. Determine where the signal receiver (focus) should be placed and compute its height above the vertex.
L4 Analyse
For \(x^2=8y\), \(a=2\). Focus at \((0,2)\) — 2 m directly above the vertex, exactly at rim-height.
Q3. The team proposes widening the rim to 10 m while keeping the depth 2 m. Evaluate whether the focus moves up or down and by how much.
L5 Evaluate
New rim at \((\pm 5,2)\): \(25=8a\Rightarrow a=\tfrac{25}{8}=3.125\) m. Focus moves up by \(3.125-2=1.125\) m. A wider, shallower dish has a higher focus.
Q4. Design a parabolic dish of rim-width 12 m with the focus placed exactly 3 m above the vertex. State the equation and required depth.
L6 Create
Focus height = \(a=3\). Equation: \(x^2=12y\). Depth at rim: \(x=6\Rightarrow y=\tfrac{36}{12}=3\) m. Dish depth = 3 m.
Assertion–Reason Questions
Assertion (A): Every point on the parabola \(y^2=4ax\) is equidistant from the focus \((a,0)\) and the directrix \(x=-a\).
Reason (R): A parabola is defined as the locus of points equidistant from a fixed focus and a fixed directrix.
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. The defining property is precisely what R states.
Assertion (A): The latus rectum of \(y^2=8x\) has length 8.
Reason (R): For the parabola \(y^2=4ax\), the latus rectum has length \(4a\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. \(4a=8\), length \(=4a=8\); R supplies the general formula.
Assertion (A): The parabola \(x^2=-16y\) opens to the left.
Reason (R): A negative sign in a parabola equation means the parabola opens in the negative axis direction.
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: D. A is false — \(x^2=-16y\) opens downward (negative \(y\)), not left. R is correct in spirit: negative sign flips to the negative axis direction, but the axis here is \(y\), not \(x\).

Frequently Asked Questions

What is the standard equation of a parabola?
The four standard forms with vertex at origin are y squared = 4 a x, y squared = -4 a x, x squared = 4 a y and x squared = -4 a y, where a > 0.
What is the focus of the parabola y squared = 4 a x?
The focus is at (a, 0) and the directrix is the vertical line x = -a. The axis is the x-axis and the vertex is the origin.
What is the latus rectum of a parabola?
The latus rectum is the chord through the focus perpendicular to the axis. Its length equals 4 a for the parabola y squared = 4 a x.
What is the axis of a parabola?
The axis of a parabola is the line through the focus perpendicular to the directrix. The parabola is symmetric about this axis.
Why does a satellite dish use a parabolic shape?
Parallel signals striking a parabolic reflector all bounce to the focus. This makes the parabola ideal for antennas and car headlights (source at focus gives parallel beam).
Does the parabola close up like an ellipse?
No. A parabola has a single open branch and extends to infinity along its axis; it does not close up like an ellipse or a circle.
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