Parabola

Focal Chord and Focal Properties

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Focal Chord and Focal Properties

Focal Chord and Focal Properties: Your JEE Main Parabola Power-Up!

Hey JEE aspirants! Get ready to dive deep into the focal chord and its fascinating properties within the world of parabolas. This isn't just about memorizing formulas; it's about understanding the geometry and leveraging that understanding to solve problems quickly and accurately. Mastering these concepts is crucial for acing coordinate geometry questions in JEE Main. Let's unlock the secrets!

What is a Focal Chord?

Imagine a line passing through the focus of a parabola. If that line intersects the parabola at two distinct points, then that line segment joining those two points is called a focal chord. Simple as that! The focus is the heart of the parabola, and the focal chord gives us some special relationships.

Endpoints of a Focal Chord: The t₁t₂ = -1 Connection

Let's consider the standard parabola equation: y2=4axy^2 = 4ax. Any point on this parabola can be represented as (at2,2at)(at^2, 2at), where tt is a parameter. Now, let's say a focal chord intersects the parabola at points P(at12,2at1)P(at_1^2, 2at_1) and Q(at22,2at2)Q(at_2^2, 2at_2). Here's the magic:

The endpoints of a focal chord satisfy the relation:

t1t2=1t_1 t_2 = -1

Geometric Intuition: This relationship tells us that if you know the parameter of one endpoint of a focal chord, you automatically know the parameter of the other endpoint! It's a direct link. Think of it as a symmetry property around the focus.

Why is t1t2=1t_1 t_2 = -1? The slope of the line joining P(at12,2at1)P(at_1^2, 2at_1) and Q(at22,2at2)Q(at_2^2, 2at_2) is 2a(t2t1)a(t22t12)=2t1+t2\frac{2a(t_2 - t_1)}{a(t_2^2 - t_1^2)} = \frac{2}{t_1 + t_2}. For the line to be a focal chord, it has to pass through the focus (a,0)(a, 0). Applying the slope condition and solving yields t1t2=1t_1t_2 = -1.

Focal Distance: SP = |x + a| - Measuring from the Heart

The focal distance of a point P(x,y)P(x, y) on the parabola is its distance from the focus S(a,0)S(a, 0). It's denoted as SPSP. Here's a super useful formula:

Focal Distance:

SP=x+aSP = |x + a|

Geometric Intuition: The focal distance is simply the distance of the point P from the directrix x=ax = -a. In other words, it's just how far the point is horizontally from the directrix. This connects the focus and directrix in a fundamental way.

Why is SP=x+aSP = |x + a|? By definition of the parabola, the distance of any point on the parabola from the focus is equal to its distance from the directrix. The distance of a point (x,y)(x, y) from the directrix x+a=0x + a = 0 is x+a|x + a|.

Harmonic Mean Property of the Focal Chord: A Balancing Act

Let PP and QQ be the endpoints of a focal chord. The semi-latus rectum (half the length of the latus rectum) is the harmonic mean between the segments SPSP and SQSQ. Remember, harmonic mean relates reciprocals.

Harmonic Mean Property:

1SP+1SQ=1a\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a}

Geometric Intuition: This property tells us that the lengths of the two focal chord segments, when combined in a specific reciprocal way, give us a constant value directly related to the parabola's parameter, aa. It's a beautiful balance.

Why is 1SP+1SQ=1a\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a}? Let P(at12,2at1)P(at_1^2, 2at_1) and Q(at22,2at2)Q(at_2^2, 2at_2) be the endpoints. Then SP=a(1+t12)SP = a(1 + t_1^2) and SQ=a(1+t22)SQ = a(1 + t_2^2). Using t1t2=1t_1t_2 = -1, you can show that 1SP+1SQ=1a\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a}.

Length of the Focal Chord: Unleash the Power of the Parameter

We can express the length of the focal chord PQ in terms of the parameter t1t_1 (or t2t_2).

Length of Focal Chord:

a(t1t2)2=a(t+1t)2a(t_1 - t_2)^2 = a\left(t + \frac{1}{t}\right)^2

Since t1t2=1t_1t_2 = -1, we can write t2=1t1t_2 = -\frac{1}{t_1}. For simplicity, let t1=tt_1 = t, which makes t2=1tt_2 = -\frac{1}{t}. Substituting into the formula a(t1t2)2a(t_1 - t_2)^2 gives us a(t+1t)2a\left(t + \frac{1}{t}\right)^2.

Length of Focal Chord at Angle θ:

Length=4asin2θ\text{Length} = \frac{4a}{\sin^2\theta}

Where θ\theta is the angle the focal chord makes with the axis of the parabola.

Tips for Conquering Focal Chord Problems

  • Draw diagrams! Coordinate geometry thrives on visualization. Sketch the parabola, the focus, and the focal chord to get a feel for the problem.
  • Use t1t2=1t_1t_2 = -1 strategically. This is your key to linking the endpoints.
  • Think about symmetry. The parabola is symmetric about its axis. How does this affect the focal chord?
  • Relate everything to 'a'. The parameter 'a' is the fundamental property of the parabola. Express everything in terms of 'a' whenever possible.

Common Mistakes to Dodge

  • Forgetting the absolute value in SP=x+aSP = |x + a|. Distance is always positive!
  • Mixing up t1t_1 and t2t_2. Remember, they're related by t1t2=1t_1t_2 = -1.
  • Ignoring the geometric interpretation. Don't just memorize formulas; understand what they represent.
  • Assuming the parabola is always y2=4axy^2 = 4ax. Be careful about transformations and shifts.

JEE-Specific Tricks

  • Quick Elimination: If a problem involves focal chords, quickly check if the given options satisfy t1t2=1t_1t_2 = -1. You can eliminate incorrect options instantly.
  • Parametric Form Dominance: Leverage the parametric form (at2,2at)(at^2, 2at) extensively. It simplifies calculations and reveals hidden relationships.
  • Latus Rectum as a Special Case: Remember the latus rectum is a special focal chord perpendicular to the axis. This can provide a quick check or a starting point for some problems.

So there you have it! Focal chords and their properties, demystified and ready for your JEE Main assault. Practice these concepts, visualize the geometry, and you'll be solving parabola problems like a pro!