JEE Main 2025 (January)

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Animated Solution for Mathematics - Conic Sections: Let $P (4, 43)$ be a point on the parabola $y^{2} = 4 a x$ and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:

Q: Do I really need to find the coordinates of $Q$ using the parameter $t_2$? Can't I just use the focal chord property $x_1 x_2 = a^2$?

You absolutely can! The properties $x_1 x_2 = a^2$ and $y_1 y_2 = -4a^2$ are powerful shortcuts for focal chords. Using them can save you time, but deriving $t_2$ is a fundamental method that works even if you forget the specific coordinate shortcuts. Both paths lead to the same beautiful result.

Q: Why is the height of the trapezium $MN$ just the difference in $y$-coordinates?

Because $M$ and $N$ both lie on the same vertical line $x = -3$. Since they share the same $x$-coordinate, the distance between them is simply the absolute difference of their $y$-coordinates. Visualizing the geometry on the Cartesian plane makes this clear: $MN = |y_P - y_Q|$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The directrix of the parabola

y^{2} = 4 a x

is a vertical line

x = - a

. Since

M

and

N

are feet of perpendiculars from

P

and

Q

to this vertical line, the segments

P M

and

QN

are horizontal. What does this imply about the quadrilateral

P QM N

Hint 2

Hint 2: Recall the parametric coordinates of a point on the parabola

y^{2} = 4 a x

are

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

. For a focal chord

P Q

, the parameters

t_{1}

and

t_{2}

of the endpoints

P

and

Q

satisfy a very specific relationship:

t_{1} t_{2} = - 1

. Use this to find the coordinates of

Q

Hint 3

Hint 3: Once you have the coordinates of

P

and

Q

, the lengths of the parallel sides

P M

and

QN

are simply the horizontal distances from the points to the directrix

x = - a

. The height of the trapezium

M N

is the vertical distance between

P

and

Q

. The area of a trapezium is

\frac{1}{2} (sum of parallel sides) \times height

Visualized Solution

Point on the Parabola

Point $P (4, 43)$ lies on the parabola $y^{2} = 4 a x$ .

Finding Parameter $a$

Substitute $x = 4, y = 43$ into $y^{2} = 4 a x$ :
$(43)^{2} = 4 a (4)$

Equation of the Parabola

$48 = 16 a ⟹ a = 3$
Parabola equation: $y^{2} = 12 x$

Focus and Directrix

For $y^{2} = 12 x$ , $a = 3$ .
Focus $S (a, 0) ⟹ S (3, 0)$
Directrix $x = - a ⟹ x = - 3$

Parametric Form of $P$

Let $P$ be $(a t_{1}^{2}, 2 a t_{1})$ .
Since $a = 3$ , $P = (3 t_{1}^{2}, 6 t_{1})$ .
Equating y-coordinates with $P (4, 43)$ :
$6 t_{1} = 43$

Parameter $t_{1}$ for $P$

$t_{1} = \frac{4 3}{6} = \frac{2}{3}$

Focal Chord Property

For a focal chord $P Q$ , the parameters satisfy $t_{1} t_{2} = - 1$ .
Substitute $t_{1} = \frac{2}{3}$ :
$(\frac{2}{3}) t_{2} = - 1$

Coordinates of $Q$

$t_{2} = - \frac{3}{2}$
$Q = (3 t_{2}^{2}, 6 t_{2}) = (3 (- \frac{3}{2})^{2}, 6 (- \frac{3}{2}))$
$Q = (\frac{9}{4}, - 33)$

Feet of Perpendiculars $M$ and $N$

Draw perpendiculars from $P$ and $Q$ to the directrix $x = - 3$ .
$M = (- 3, 43)$
$N = (- 3, - 33)$

Quadrilateral $P QM N$

$P M$ and $QN$ are perpendicular to the same line $x = - 3$ .
Therefore, $P M$ is parallel to $QN$ .
Quadrilateral $P QM N$ is a trapezium.

Dimensions of Trapezium

Parallel sides:
$P M = 4 - (- 3) = 7$
$QN = \frac{9}{4} - (- 3) = \frac{21}{4}$
Height $M N = 43 - (- 33) = 73$

Area Setup

Area of trapezium = $\frac{1}{2} (P M + QN) \times M N$
Substitute the lengths:
Area = $\frac{1}{2} (7 + \frac{21}{4}) \times 73$

Final Area

Simplify the sum: $7 + \frac{21}{4} = \frac{49}{4}$
Area = $\frac{1}{2} (\frac{49}{4}) \times 73$
Final Area = $\frac{343 3}{8}$

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

Solution Diagram

Analyzing the Setup

We are given a point

P (4, 43)

on the parabola

y^{2} = 4 a x

. To identify the specific parabola, we substitute the coordinates of

P

into the equation:

(43)^{2} = 4 a (4)

This simplifies to

48 = 16 a

, which yields

a = 3

. Therefore, the equation of our parabola is

y^{2} = 12 x

With

a = 3

, the focus

S

is located at

(3, 0)

and the directrix is the vertical line

x = - 3

The Focal Chord Mystery

A focal chord is a line segment passing through the focus

S

. The parametric coordinates of any point on the parabola

y^{2} = 12 x

are given by

(3 t^{2}, 6 t)

For point

P

, we equate the

y

-coordinate

6 t_{1} = 43

, which gives the parameter

t_{1} = \frac{2}{3}

For any focal chord, the product of the parameters of its endpoints is always

- 1

. Thus,

t_{1} t_{2} = - 1

implies:

t_{2} = - \frac{1}{t _{1}} = - \frac{3}{2}

Using

t_{2}

, we find the coordinates of point

Q

Q = 3 (- \frac{3}{2})^{2}, 6 (- \frac{3}{2}) = (\frac{9}{4}, - 33)

The Geometry of the Trapezium

We draw perpendiculars from

P

and

Q

to the directrix

x = - 3

. These perpendiculars meet the directrix at

M (- 3, 43)

and

N (- 3, - 33)

Since

P M

and

QN

are both perpendicular to the same vertical line, they are horizontal and parallel. This confirms that

P QM N

is a trapezium.

The lengths of the parallel sides are the horizontal distances:

P M = 4 - (- 3) = 7

QN = \frac{9}{4} - (- 3) = \frac{21}{4}

The height of the trapezium is the vertical distance between

M

and

N

M N = 43 - (- 33) = 73

Final Calculation

We apply the area formula for a trapezium:

Area = \frac{1}{2} (P M + QN) \times M N

Substituting our values:

Area = \frac{1}{2} (7 + \frac{21}{4}) \times 73

Simplifying the sum inside the parentheses:

Area = \frac{1}{2} (\frac{28 + 21}{4}) \times 73 = \frac{1}{2} (\frac{49}{4}) \times 73

The final area of the trapezium is:

Area = \frac{343 3}{8}

Common Traps & FAQs

Wait, why is

P QM N

a trapezium? I thought it might be a rectangle or something else.

That is a fantastic question! A trapezium is defined as a quadrilateral with at least one pair of parallel sides. Since

P M

and

QN

are both perpendicular to the same vertical line (the directrix

x = - 3

), they are both horizontal segments. Because they are both horizontal, they are parallel to each other. That is all we need to classify it as a trapezium!

Do I really need to find the coordinates of

Q

using the parameter

t_{2}

? Can't I just use the focal chord property

x_{1} x_{2} = a^{2}

You absolutely can! The properties

x_{1} x_{2} = a^{2}

and

y_{1} y_{2} = - 4 a^{2}

are powerful shortcuts for focal chords. Using them can save you time, but deriving

t_{2}

is a fundamental method that works even if you forget the specific coordinate shortcuts. Both paths lead to the same beautiful result.

Why is the height of the trapezium

M N

just the difference in

y

-coordinates?

Because

M

and

N

both lie on the same vertical line

x = - 3

. Since they share the same

x

-coordinate, the distance between them is simply the absolute difference of their

y

-coordinates. Visualizing the geometry on the Cartesian plane makes this clear:

M N = ∣ y_{P} - y_{Q} ∣

Select Answer:

Visualized Solution

Point on the Parabola

Finding Parameter $a$

Equation of the Parabola

Focus and Directrix

Parametric Form of $P$

Parameter $t_{1}$ for $P$

Focal Chord Property

Coordinates of $Q$

Feet of Perpendiculars $M$ and $N$

Quadrilateral $P QM N$

Dimensions of Trapezium

Area Setup

Final Area

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

Analyzing the Setup

The Focal Chord Mystery

The Geometry of the Trapezium

Final Calculation

Similar Questions

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is

Comprehension Passage

The orthocentre of the triangle $F_{1} M N$ is

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Tangents drawn from the point (-8, 0) to the parabola $y^{2} = 8 x$ touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :-

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Select Answer:

Visualized Solution

Point on the Parabola

Finding Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Equation of the Parabola

Focus and Directrix

Parametric Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Focal Chord Property

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Feet of Perpendiculars .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Quadrilateral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQMN

Dimensions of Trapezium

Area Setup

Final Area

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

Analyzing the Setup

The Focal Chord Mystery

The Geometry of the Trapezium

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the focus of the parabola y2=8x and let PQ be the common chord of the circle x2+y2−2x−4y=0 and the given parabola. The area of the triangle PQS is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The orthocentre of the triangle F1​MN is

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Quadrilateral $P QM N$

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is

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