If the chord joining the points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ on the parabola $y^2 = 12x$ subtends a right angle at the vertex of the parabola, then $x_1x_2 – y_1y_2$ is equal to
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Solution Steps
1
Parametric form of parabola
Given $y^2 = 12x \;\Rightarrow\; 4a = 12 \;\Rightarrow\; a = 3$.
Standard parametric: $x = at^2 = 3t^2$, $y = 2at = 6t$.
Let $P_1 \equiv (3t_1^2,\; 6t_1)$, $P_2 \equiv (3t_2^2,\; 6t_2)$.
2
Right angle at vertex
Vertex $V \equiv (0,0)$. Vectors $\overrightarrow{VP_1} = (3t_1^2,\;6t_1)$, $\overrightarrow{VP_2} = (3t_2^2,\;6t_2)$.
Condition for right angle: $\overrightarrow{VP_1} \cdot \overrightarrow{VP_2} = 0$.
$(3t_1^2)(3t_2^2) + (6t_1)(6t_2) = 0$.
$9t_1^2 t_2^2 + 36 t_1 t_2 = 0$.
3
Simplify the condition
Factor $9t_1t_2$: $9t_1t_2\,(t_1t_2 + 4) = 0$.
Chord endpoints are distinct and not at vertex, so $t_1\neq0,\;t_2\neq0$, hence $t_1t_2 + 4 = 0$.
Thus $\boxed{t_1t_2 = -4}$.
4
Compute $x_1x_2$ and $y_1y_2$
$x_1x_2 = (3t_1^2)(3t_2^2) = 9(t_1t_2)^2$.
$y_1y_2 = (6t_1)(6t_2) = 36(t_1t_2)$.
Substitute $t_1t_2 = -4$:
$x_1x_2 = 9 \times 16 = 144$, $y_1y_2 = 36 \times (-4) = -144$.
5
Final expression
$x_1x_2 – y_1y_2 = 144 – (-144) = 144 + 144 = 288$.
Hence the value is $\mathbf{288}$.
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Key condition
$t_1t_2 = -4$ for any chord of $y^2=4ax$ subtending right angle at vertex. Then $x_1x_2 = a^2(t_1t_2)^2 = 16a^2$ and $y_1y_2 = 4a^2(t_1t_2) = -16a^2$.
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Parabola Theory
1. Parametric representation of $y^2 = 4ax$
Any point on the parabola $y^2 = 4ax$ can be written as $(at^2, 2at)$ where $t \in \mathbb{R}$ is a parameter. For $y^2 = 12x$, we have $4a = 12$ ⇒ $a=3$, so points become $(3t^2, 6t)$. This parametric form is extremely useful for problems involving chords, focal properties, and tangents because it reduces algebraic complexity.
2. Chord joining two parametric points
If $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$, the equation of chord $PQ$ is $y(t_1+t_2) = 2x + 2a t_1t_2$. Many conditions like “subtending right angle at vertex”, “passing through a fixed point”, or “focal chord” translate directly into simple relations between $t_1$ and $t_2$, making parametric approach the most efficient in JEE problems.
3. Right angle at the vertex condition
Vertex $V(0,0)$ is the simplest point. For chord $PQ$ to subtend a right angle at $V$, vectors $VP$ and $VQ$ are perpendicular: $(at_1^2)(at_2^2) + (2at_1)(2at_2)=0$ ⇒ $a^2t_1^2t_2^2+4a^2t_1t_2=0$ ⇒ $a^2t_1t_2(t_1t_2+4)=0$. Since $t_1,t_2 \neq 0$ for a non‑degenerate chord, $t_1t_2 = -4$. Notice that the value $-4$ is independent of $a$! But $a$ reappears when computing expressions like $x_1x_2$ or $y_1y_2$.
4. Symmetric expressions: $x_1x_2 \pm y_1y_2$
For $y^2=4ax$, $x_1x_2 = a^2(t_1t_2)^2$ and $y_1y_2 = 4a^2t_1t_2$. Using $t_1t_2 = -4$ we get $x_1x_2 = a^2(16)=16a^2$, $y_1y_2 = 4a^2(-4) = -16a^2$. Therefore $x_1x_2 – y_1y_2 = 16a^2 – (-16a^2) = 32a^2$. For our $a=3$, $32 \times 9 = 288$. This matches the answer. Memorizing $t_1t_2 = -4$ helps solve many variants quickly.
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FAQs
Q
Why did we take parametric coordinates?
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Because the condition “subtends right angle at vertex” becomes a simple dot product between vectors expressed in terms of parameters t₁ and t₂, avoiding messy algebra with x,y coordinates.
Q
Could t₁ or t₂ be zero?
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No. If t = 0, the point is the vertex itself (0,0). Then the chord degenerates to a single point, not a chord.
Q
What if the parabola was y² = –12x?
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Then a = –3, parametric: (–3t², 6t). Dot product condition leads to t₁t₂ = 4. The expression x₁x₂ – y₁y₂ would be 9(16) – 36(4) = 144 – 144 = 0.
Q
Is the answer 288 independent of which chord?
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Yes, for any chord of y² = 12x that subtends a right angle at vertex, x₁x₂ – y₁y₂ always equals 288.
Q
What does “subtends a right angle at the vertex” mean geometrically?
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It means the lines joining the vertex to the two endpoints of the chord are perpendicular. ∠P₁VP₂ = 90°.
Q
What if we had used slope form?
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We could write chord equation and homogenize with parabola, but parametric method is shorter.
Q
What is a chord in parabola?
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A line segment joining any two distinct points on the parabola.
Q
What if the question was for x² = 12y?
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Then parametric: (6t, 3t²). Condition would change completely. Always derive from dot product.
Q
Why is the answer 288 and not 144?
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Because x₁x₂ = 144 and y₁y₂ = –144, so x₁x₂ – y₁y₂ = 144 – (–144) = 288.
Q
What is the use of such problems?
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They test parametric representation, condition for perpendicularity, and algebraic manipulation — all essential for JEE.
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