JEE Main 2025 April

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Animated Solution for Mathematics - Conic Sections: Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $2 a$ and $2 b$ , respectively, and one focus and the corresponding directrix of this hyperbola be $(- 5, 0)$ and $5 x + 9 = 0$ , respectively. If the product of the focal distances of a point $(α, 25)$ on the hyperbola is $p$ , then $4 p$ is equal to

Q: How do I know for sure that the hyperbola is horizontal?

Look at the focus $(-5, 0)$. It lies on the x-axis. The directrix $x = -\frac{9}{5}$ is a vertical line perpendicular to the x-axis. This specific configuration—a focus on the x-axis and a vertical directrix—is the hallmark of a standard horizontal hyperbola centered at the origin.

Q: Why is the product of focal distances equal to $|e^2x^2 - a^2|$?

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize the coordinate plane. You have a focus at

(- 5, 0)

and a directrix

x = - \frac{9}{5}

. The distance from the center to the focus is

a e

, and the distance from the center to the directrix is

\frac{a}{e}

. Can you see how these two simple relations allow you to isolate

a

and

e

Hint 2

Hint 2: Do not fall into the trap of calculating the distance from the point to the focus using the standard coordinate distance formula. That is a path to algebraic misery. Instead, recall the elegant property that the focal distances of a point

(x, y)

on a hyperbola are

∣ e x - a ∣

and

∣ e x + a ∣

Hint 3

Hint 3: Start by solving the system

a e = 5

and

\frac{a}{e} = \frac{9}{5}

. Once you have

a

and

e

, find

b^{2}

using the fundamental relation

b^{2} = a^{2} (e^{2} - 1)

. Finally, substitute the point

(α, 25)

into the hyperbola equation to find

α^{2}

, and plug everything into the product formula

∣ e^{2} x^{2} - a^{2} ∣

Visualized Solution

Visualizing the Hyperbola Setup

Standard Hyperbola: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Given Focus: $S (- 5, 0)$
Given Directrix: $5 x + 9 = 0 ⟹ x = - \frac{9}{5}$

Relating Geometry to Parameters

Distance of focus from center: $a e = 5$
Distance of directrix from center: $\frac{a}{e} = \frac{9}{5}$

Solving for $a^{2}$

Multiply the two relations to eliminate $e$ :
$(a e) \cdot (\frac{a}{e}) = 5 \cdot \frac{9}{5}$
$a^{2} = 9 ⟹ a = 3$

Finding Eccentricity $e$

Substitute $a = 3$ into $a e = 5$ :
$3 e = 5 ⟹ e = \frac{5}{3}$

Calculating $b^{2}$

Fundamental relation: $b^{2} = a^{2} (e^{2} - 1)$
Substitute values: $b^{2} = 9 ((\frac{5}{3})^{2} - 1)$
$b^{2} = 9 (\frac{25}{9} - 1) = 9 (\frac{16}{9}) = 16$

Introducing Point $P$

Equation of Hyperbola: $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$
Point $P (α, 25)$ lies on this hyperbola.

Substituting Point $P$

Substitute $x = α$ and $y = 25$ into the equation:
$\frac{α ^{2}}{9} - \frac{( 2 5 ) ^{2}}{16} = 1$

Solving for $α^{2}$

Calculate the $y$ term: $(25)^{2} = 20$
$\frac{α ^{2}}{9} - \frac{20}{16} = 1 ⟹ \frac{α ^{2}}{9} - \frac{5}{4} = 1$
$\frac{α ^{2}}{9} = 1 + \frac{5}{4} = \frac{9}{4} ⟹ α^{2} = \frac{81}{4}$

The Focal Distance Property

Focal distances of $P (x, y)$ are $∣ e x - a ∣$ and $∣ e x + a ∣$ .
Their product $p = ∣ (e x - a) (e x + a) ∣ = ∣ e^{2} x^{2} - a^{2} ∣$

Substituting into the Product Formula

We know: $e^{2} = \frac{25}{9}$ , $x^{2} = α^{2} = \frac{81}{4}$ , $a^{2} = 9$
Substitute into $p$ : $p = ∣ \frac{25}{9} \cdot \frac{81}{4} - 9∣$

Calculating $p$

$p = ∣ \frac{25 \cdot 9}{4} - 9∣ = ∣ \frac{225}{4} - 9∣$
$p = ∣ \frac{225 - 36}{4} ∣ = \frac{189}{4}$

Final Answer: $4 p$

The question asks for $4 p$ .
$4 p = 4 \cdot \frac{189}{4}$
$4 p = 189$

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

Solution Diagram

Analyzing the Setup

Imagine you are standing on the Cartesian plane. You are given a focus at

S (- 5, 0)

and a directrix at

x = - \frac{9}{5}

. The hyperbola is centered at the origin, following the standard form:

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

The geometry of the hyperbola is defined by its parameters

a

and

e

. The distance from the center to the focus is

a e = 5

, and the distance from the center to the directrix is

\frac{a}{e} = \frac{9}{5}

We have two variables and two equations. If we multiply these two relations, the eccentricity

e

vanishes entirely:

(a e) \cdot (\frac{a}{e}) = 5 \cdot \frac{9}{5}

a^{2} = 9 \Rightarrow a = 3

Constructing the Hyperbola

With

a = 3

determined, finding the eccentricity

e

is trivial. Substituting back into

a e = 5

, we find

3 e = 5

, so

e = \frac{5}{3}

Note that

e > 1

, which is the defining characteristic of a hyperbola. Now, we calculate

b^{2}

using the fundamental relation

b^{2} = a^{2} (e^{2} - 1)

b^{2} = 9 ((\frac{5}{3})^{2} - 1) = 9 (\frac{25}{9} - 1) = 9 (\frac{16}{9}) = 16

Our hyperbola is now fully defined by the equation:

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

The Point $P$ and the Focal Property

Consider a point

P (α, 25)

resting on this curve. Because it lies on the hyperbola, it must satisfy the equation. Substituting

x = α

and

y = 25

\frac{α ^{2}}{9} - \frac{( 2 5 ) ^{2}}{16} = 1

Since

(25)^{2} = 20

, the equation becomes:

\frac{α ^{2}}{9} - \frac{20}{16} = 1 \Rightarrow \frac{α ^{2}}{9} - \frac{5}{4} = 1

Solving for

α^{2}

, we find:

\frac{α ^{2}}{9} = \frac{9}{4} \Rightarrow α^{2} = \frac{81}{4}

The Grand Finale

The focal distance

p

from a point on the hyperbola can be expressed using the property

p = ∣ e x \pm a ∣

. However, for the product of focal distances, we utilize the property

p = ∣ e^{2} x^{2} - a^{2} ∣

Substituting our known values

e^{2} = \frac{25}{9}

x^{2} = \frac{81}{4}

, and

a^{2} = 9

p = \frac{25}{9} \cdot \frac{81}{4} - 9 = \frac{225}{4} - 9 = \frac{225 - 36}{4} = \frac{189}{4}

The question asks for the value of

4 p

4 p = 4 \cdot \frac{189}{4} = 189

Common Traps & FAQs

Wait, why can't we just use the distance formula for the focal distances? It seems more direct.

You absolutely could, but it would be a nightmare of square roots and radicals. The JEE Advanced examiner tests your knowledge of properties, not just your ability to grind through algebra. The property

∣ e x \pm a ∣

is a powerful tool designed to simplify exactly this type of problem, turning a complex radical expression into a clean algebraic one.

How do I know for sure that the hyperbola is horizontal?

Look at the focus

(- 5, 0)

. It lies on the x-axis. The directrix

x = - \frac{9}{5}

is a vertical line perpendicular to the x-axis. This specific configuration—a focus on the x-axis and a vertical directrix—is the hallmark of a standard horizontal hyperbola centered at the origin.

Why is the product of focal distances equal to

∣ e^{2} x^{2} - a^{2} ∣

Great question! The focal distances are

d_{1} = ∣ e x - a ∣

and

d_{2} = ∣ e x + a ∣

. When you multiply them, you get

∣ (e x - a) (e x + a) ∣

. Using the difference of squares identity, this simplifies to

∣ e^{2} x^{2} - a^{2} ∣

. It is a beautiful example of how algebraic identities simplify geometric complexity.

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola Setup

Relating Geometry to Parameters

Solving for $a^{2}$

Finding Eccentricity $e$

Calculating $b^{2}$

Introducing Point $P$

Substituting Point $P$

Solving for $α^{2}$

The Focal Distance Property

Substituting into the Product Formula

Calculating $p$

Final Answer: $4 p$

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

Analyzing the Setup

Constructing the Hyperbola

The Point $P$ and the Focal Property

The Grand Finale

Similar Questions

If $5 x + 9 = 0$ is the directrix of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is :

Let a hyperbola passes through the focus of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ . The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then

Let P be a point on the hyperbola $H : \frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ , in the first quadrant such that the area of triangle formed by P and the two foci of H is $213$ . Then, the square of the distance of P from the origin is

If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 23)$ is $5 x = 45$ and its eccentricity is $e$ , then :

Let the foci and length of the latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be $(\pm 5, 0)$ and $50$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x ^{2}}{b ^{2}} - \frac{y ^{2}}{a ^{2} b ^{2}} = 1$ equals

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola Setup

Relating Geometry to Parameters

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

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

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

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2

The Focal Distance Property

Substituting into the Product Formula

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

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4p

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

Analyzing the Setup

Constructing the Hyperbola

The Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and the Focal Property

The Grand Finale

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 5x+9=0 is the directrix of the hyperbola 16x2−9y2=144, then its corresponding focus is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a directrix of a hyperbola centred at the origin and passing through the point (4,−23​) is 5x=45​ and its eccentricity is e, then :

Solving for $a^{2}$

Finding Eccentricity $e$

Calculating $b^{2}$

Introducing Point $P$

Substituting Point $P$

Solving for $α^{2}$

Calculating $p$

Final Answer: $4 p$

The Point $P$ and the Focal Property

If $5 x + 9 = 0$ is the directrix of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is :

If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 23)$ is $5 x = 45$ and its eccentricity is $e$ , then :