JEE Main 2026 (22 January Shift 1)

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Animated Solution for Mathematics - Conic Sections: If the chord joining the points $P_{1} (x_{1}, y_{1})$ and $P_{2} (x_{2}, y_{2})$ on the parabola $y^{2} = 12 x$ subtends a right angle at the vertex of the parabola, then $x_{1} x_{2} - y_{1} y_{2}$ is equal to

Q: Wait, why do we use parametric coordinates instead of just $(x_1, y_1)$ and $(x_2, y_2)$?

Great question! While you could use the standard $(x, y)$ coordinates, you would end up with a system of equations involving square roots or complex algebraic manipulation. Parametric coordinates $(at^2, 2at)$ reduce the entire problem to a single variable $t$ for each point, making the perpendicularity condition and the final calculation much more elegant and less prone to error.

Q: Is the relationship $t_1t_2 = -4$ always true for any parabola?

Not quite! The relationship $t_1t_2 = -4$ is specific to the parabola $y^2 = 4ax$ when the chord subtends a right angle at the vertex. If the parabola were $y^2 = 4ax$, the condition $m_1m_2 = -1$ leads to $(\frac{2}{t_1})(\frac{2}{t_2}) = -1$, which simplifies to $t_1t_2 = -4$. If the parabola equation changes, the constant on the right side will change accordingly.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the parabola

y^{2} = 12 x

and its vertex at the origin

O (0, 0)

. When a chord

P_{1} P_{2}

subtends a right angle at the vertex, the vectors

O P_{1}

and

O P_{2}

are perpendicular. This means the product of their slopes must be

- 1

Hint 2

Hint 2: The Arsenal. Utilize the parametric form of the parabola

y^{2} = 4 a x

. Since

4 a = 12

, we have

a = 3

. Any point on this parabola can be represented as

(3 t^{2}, 6 t)

. Express the slopes of

O P_{1}

and

O P_{2}

in terms of the parameters

t_{1}

and

t_{2}

Hint 3

Hint 3: The Execution. The slope of

O P_{1}

\frac{6 t _{1}}{3 t _{1}^{2}} = \frac{2}{t _{1}}

and the slope of

O P_{2}

\frac{2}{t _{2}}

. Setting their product to

- 1

gives

\frac{4}{t _{1} t _{2}} = - 1

, which implies

t_{1} t_{2} = - 4

. Now, substitute this into the expression

x_{1} x_{2} - y_{1} y_{2}

using the parametric coordinates.

Visualized Solution

Analyze the Parabola $y^{2} = 12 x$

Given Parabola: $y^{2} = 12 x$
Standard Form: $y^{2} = 4 a x$
Comparing coefficients: $4 a = 12 \Rightarrow a = 3$

Define Parametric Points $P_{1}$ and $P_{2}$

Parametric coordinates on $y^{2} = 4 a x$ are $(a t^{2}, 2 a t)$
With $a = 3$ :
$P_{1} = (3 t_{1}^{2}, 6 t_{1})$
$P_{2} = (3 t_{2}^{2}, 6 t_{2})$

The Right Angle Condition at Vertex

Vertex of the parabola: $O (0, 0)$
The chord $P_{1} P_{2}$ subtends $9 0^{\circ}$ at the vertex $O$ .
Condition: $O P_{1} ⊥ O P_{2}$

Calculate Individual Slopes

Slope of $O P_{1} (m_{1}) = \frac{6 t _{1} - 0}{3 t _{1}^{2} - 0} = \frac{2}{t _{1}}$
Slope of $O P_{2} (m_{2}) = \frac{6 t _{2} - 0}{3 t _{2}^{2} - 0} = \frac{2}{t _{2}}$

Apply Orthogonality Condition

For perpendicular lines: $m_{1} \cdot m_{2} = - 1$
Substitute slopes: $(\frac{2}{t _{1}}) \cdot (\frac{2}{t _{2}}) = - 1$

Simplify the Parameter Relation

Multiply the terms: $\frac{4}{t _{1} t _{2}} = - 1$
Rearrange to find $t_{1} t_{2}$ : $t_{1} t_{2} = - 4$

Express $x_{1} x_{2}$ in Parameters

Target expression involves $x_{1} x_{2}$ and $y_{1} y_{2}$ .
Calculate $x_{1} x_{2}$ :
$x_{1} x_{2} = (3 t_{1}^{2}) (3 t_{2}^{2}) = 9 (t_{1} t_{2})^{2}$

Express $y_{1} y_{2}$ in Parameters

Calculate $y_{1} y_{2}$ :
$y_{1} y_{2} = (6 t_{1}) (6 t_{2}) = 36 (t_{1} t_{2})$

Substitute $t_{1} t_{2} = - 4$

Recall: $t_{1} t_{2} = - 4$
$x_{1} x_{2} = 9 (- 4)^{2} = 9 (16) = 144$
$y_{1} y_{2} = 36 (- 4) = - 144$

Final Calculation

Final expression: $x_{1} x_{2} - y_{1} y_{2}$
Substitute the computed values: $144 - (- 144)$
$= 144 + 144 = 288$

00:00 / 00:00

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

We are given the parabola defined by the equation

y^{2} = 12 x

. We consider two points

P_{1}

and

P_{2}

on this curve such that the chord

P_{1} P_{2}

subtends a right angle at the vertex

O (0, 0)

By comparing the given equation with the standard form

y^{2} = 4 a x

, we identify

4 a = 12

, which yields

a = 3

The Parametric Power-up

To simplify the geometry, we utilize the parametric form of the parabola. Any point

P

on the curve can be represented as

(a t^{2}, 2 a t)

Substituting

a = 3

, the coordinates of our points are:

P_{1} = (3 t_{1}^{2}, 6 t_{1})

P_{2} = (3 t_{2}^{2}, 6 t_{2})

The Geometric Constraint

The condition that the chord

P_{1} P_{2}

subtends a right angle at the origin implies that the lines

O P_{1}

and

O P_{2}

are perpendicular. Consequently, the product of their slopes must be

- 1

The slope

m

of a line from the origin to a point

(x, y)

is given by

\frac{y}{x}

. Thus, we calculate the slopes as follows:

m_{1} = \frac{6 t _{1}}{3 t _{1}^{2}} = \frac{2}{t _{1}}

m_{2} = \frac{6 t _{2}}{3 t _{2}^{2}} = \frac{2}{t _{2}}

Applying the perpendicularity condition

m_{1} \cdot m_{2} = - 1

(\frac{2}{t _{1}}) \cdot (\frac{2}{t _{2}}) = - 1

\Rightarrow t_{1} t_{2} = - 4

The Algebraic Symphony

We now evaluate the expression

x_{1} x_{2} - y_{1} y_{2}

using our parametric relations. First, we calculate the product of the

x

-coordinates:

x_{1} x_{2} = (3 t_{1}^{2}) (3 t_{2}^{2}) = 9 (t_{1} t_{2})^{2}

x_{1} x_{2} = 9 (- 4)^{2} = 9 (16) = 144

Next, we calculate the product of the

y

-coordinates:

y_{1} y_{2} = (6 t_{1}) (6 t_{2}) = 36 (t_{1} t_{2})

y_{1} y_{2} = 36 (- 4) = - 144

Finally, we compute the required difference:

x_{1} x_{2} - y_{1} y_{2} = 144 - (- 144) = 144 + 144 = 288

The final value of the expression is 288.

Common Traps & FAQs

Wait, why do we use parametric coordinates instead of just

(x_{1}, y_{1})

and

(x_{2}, y_{2})

Great question! While you could use the standard

(x, y)

coordinates, you would end up with a system of equations involving square roots or complex algebraic manipulation. Parametric coordinates

(a t^{2}, 2 a t)

reduce the entire problem to a single variable

t

for each point, making the perpendicularity condition and the final calculation much more elegant and less prone to error.

Why is the product of the slopes equal to

- 1

This is a fundamental property of perpendicular lines in the Cartesian plane. If two lines with slopes

m_{1}

and

m_{2}

are perpendicular, their angle of intersection is

9 0^{\circ}

. The formula for the angle

θ

between two lines is

tan θ = ∣ \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} ∣

. For

θ = 9 0^{\circ}

tan θ

is undefined, which forces the denominator

1 + m_{1} m_{2}

to be zero, hence

m_{1} m_{2} = - 1

Is the relationship

t_{1} t_{2} = - 4

always true for any parabola?

Not quite! The relationship

t_{1} t_{2} = - 4

is specific to the parabola

y^{2} = 4 a x

when the chord subtends a right angle at the vertex. If the parabola were

y^{2} = 4 a x

, the condition

m_{1} m_{2} = - 1

leads to

(\frac{2}{t _{1}}) (\frac{2}{t _{2}}) = - 1

, which simplifies to

t_{1} t_{2} = - 4

. If the parabola equation changes, the constant on the right side will change accordingly.

Animated Solution for Mathematics - Conic Sections: If the chord joining the points $P_{1} (x_{1}, y_{1})$ and $P_{2} (x_{2}, y_{2})$ on the parabola $y^{2} = 12 x$ subtends a right angle at the vertex of the parabola, then $x_{1} x_{2} - y_{1} y_{2}$ is equal to

Select Answer:

Visualized Solution

Analyze the Parabola $y^{2} = 12 x$

Define Parametric Points $P_{1}$ and $P_{2}$

The Right Angle Condition at Vertex

Calculate Individual Slopes

Apply Orthogonality Condition

Simplify the Parameter Relation

Express $x_{1} x_{2}$ in Parameters

Express $y_{1} y_{2}$ in Parameters

Substitute $t_{1} t_{2} = - 4$

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Parametric Power-up

The Geometric Constraint

The Algebraic Symphony

Similar Questions

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Select Answer:

Visualized Solution

Analyze the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=12x

Define Parametric Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​

The Right Angle Condition at Vertex

Calculate Individual Slopes

Apply Orthogonality Condition

Simplify the Parameter Relation

Express .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​x2​ in Parameters

Express .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y1​y2​ in Parameters

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​t2​=−4

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Parametric Power-up

The Geometric Constraint

The Algebraic Symphony

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the chord of the parabola x2=4y having equation x−2​y+42​=0 is :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the x-intercept of a focal chord of the parabola y2=8x+4y+4 is 3, then the length of this chord is equal to

Analyze the Parabola $y^{2} = 12 x$

Define Parametric Points $P_{1}$ and $P_{2}$

Express $x_{1} x_{2}$ in Parameters

Express $y_{1} y_{2}$ in Parameters

Substitute $t_{1} t_{2} = - 4$

Let the length of the focal chord $P Q$ of the parabola $y^{2} = 12 x$ be 15 units. If the distance of $P Q$ from the origin is $p$ , then $10 p^{2}$ is equal to _______

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If one end of a focal chord of the parabola, $y^{2} = 16 x$ is at $(1, 4)$ , then the length of this focal chord is

If the x-intercept of a focal chord of the parabola $y^{2} = 8 x + 4 y + 4$ is 3, then the length of this chord is equal to