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Conic Sections appeared 76 times across 3 years — 8.8% of Mathematics. This question is from Properties of Focal Chords.

Year 2026 2025 2024 Total
Questions 22 38 16 76

Let y² = 12x the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP) (SQ) = (147)/(4). Let C be the circle described taking PQ as a diameter. If the equation of a circle C is 64x² + 64y² - α x - 64√(3)y = β, then \beta - \alpha is equal to

Numerical Answer Type:
Enter a numerical value Answer: 1328 to 1328 +4 marks

Solution & Explanation

Related Formula

Properties of focal chord parameter metrics in parabolas y² = 4ax:

t₁ · t₂ = -1

Distance to the directrix property:

SP = a(1 + t²), SQ = a(1 + (1)/(t²))
Core Logic

Given parabola y² = 12x a = 3. Focus S = (3, 0). Set up focal segments product equation:

SP · SQ = 3(1+t²) · 3(1+(1)/(t²)) = (147)/(4) 9 · ((1+t²)²)/(t²) = (147)/(4) ((1+t²)²)/(t²) = (49)/(12)

Solving for t²:

12t⁴ - 25t² + 12 = 0 t² = (3)/(4) or (4)/(3)
Step 1: Compute Endpoint Coordinate Bounds

Choosing t = - √(3)2 allows defining both chord coordinates symmetrically:

P(3t², 6t) P((9)/(4), -3√(3)) Q((3)/(t²), -(6)/(t)) Q(4, 4√(3))
Step 2: Derive Circle Equation

Write the diameter circle form equation:

(x - 4)(x - (9)/(4)) + (y - 4√(3))(y + 3√(3)) = 0 x² + y² - (25)/(4)x - √(3)y - 27 = 0

Multiply by 64 to clear the fractions and match the given equation template structure:

64x² + 64y² - 400x - 64√(3)y - 1728 = 0

Comparing directly with 64x² + 64y² - α x - 64√(3)y = β yields:

α = 400, β = 1728 β - α = 1728 - 400 = 1328
Pattern Recognition

The distance from focal chord endpoints to the focus equals their perpendicular distance to the directrix. This property connects parameter metrics to geometric lengths cleanly.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

Reference Study Guides

More Conic Sections Previous-Year Questions — Page 5

Q9 jee_main_2026_28_january_evening Ellipse Parameters and Latus Rectum
An ellipse has its center at (1,-2), one focus at (3,-2) and one vertex at (5, - 2). Then the length of its latus rectum is :
  • A. 16√(3)
  • B. 6
  • C. 4√(3)
  • D. 6√(3)

Solution

Related Formula
Latus Rectum (LR) = (2b²)/(a) = 2a(1-e²)
Core Logic

From the given coordinates on the major axis (y = -2): Center C(1, -2), Focus F₁(3, -2), Vertex A₁(5, -2). Distance from center to vertex, CA₁ = a = 5 - 1 = 4. Distance from center to focus, CF₁ = ae = 3 - 1 = 2.

Ellipse dimensions mapped to coordinates
Ellipse dimensions mapped to coordinates

Execution

Calculate eccentricity e:

ae = 2 ⇒ 4e = 2 ⇒ e = (1)/(2)

Use alternate formula for Latus Rectum:

LR = 2e((a)/(e) - ae) or directly 2a(1-e²) LR = 2(4)(1 - (1)/(4)) = 8 × (3)/(4) = 6
Pattern Recognition

Aligning focus, center, and vertex along a constant y-axis implies a standard shifted ellipse where absolute differences in x-coordinates yield standard parameters (a and ae) directly.

Chapter Mix

Class 11 Maths: Conic Sections

Q10 jee_main_2026_28_january_evening Confocal Ellipse and Hyperbola
Let the ellipse E: x²144 + y²169 = 1 and the hyperbola H: x²16 - y²λ² = -1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, then the value of 24(e + L) is:
  • A. 296
  • B. 126
  • C. 148
  • D. 67

Solution

Related Formula
e = √(1 - (a²)/(b²)) (for vertical ellipse) e = √(1 + (a²)/(b²)) (for conjugate hyperbola)
Core Logic

For Ellipse E: (x²)/(144) + (y²)/(169) = 1 a² = 144, b² = 169. Since b > a, the major axis is along the y-axis. Eccentricity e' = √(1 - (144)/(169)) = √((25)/(169)) = (5)/(13). Foci of ellipse = (0, ± be') = (0, ± 13 × (5)/(13)) = (0, ± 5).

Execution

For Hyperbola H: (y²)/(λ²) - (x²)/(16) = 1 Foci of conjugate hyperbola are (0, ± λ e). Equating foci: λ e = 5.

e = √(1 + (16)/(λ²)) λ √(1 + (16)/(λ²)) = 5 ⇒ λ² + 16 = 25 ⇒ λ² = 9 ⇒ λ = 3

Eccentricity of hyperbola, e = (5)/(3). Length of latus rectum of hyperbola, L = (2(16))/(λ) = (32)/(3).

Calculate 24(e + L):

24(e + L) = 24[(5)/(3) + (32)/(3)] = 24((37)/(3)) = 8 × 37 = 296
Pattern Recognition

Confocal conics usually align along the same major axis. Notice the -1 on the RHS of the hyperbola equation indicates a conjugate hyperbola orienting it vertically to match the b>a ellipse.

Chapter Mix

Class 11 Maths: Conic Sections

Q55 jee_main_2025_02_april_evening Ellipse
If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :
  • A. 4√(17)
  • B. √(3)16
  • C. 3√(19)
  • D. √(5)7

Solution

Related Formula
Length of minor axis = 2b Distance between foci = 2ae Eccentricity: e = √(1 - (b²)/(a²))
Core Logic

We set up an algebraic equation relating b, a, and e from the given geometric condition, then substitute it into the eccentricity identity.

Step 1: Set up the geometric relation

Given that 2b = (1)/(4) (2ae):

b = (ae)/(4) (b)/(a) = (e)/(4)

Square both sides:

(b²)/(a²) = (e²)/(16)
Step 2: Solve for eccentricity

Using the eccentricity relation:

e² = 1 - (b²)/(a²) e² = 1 - (e²)/(16) e² (1 + (1)/(16)) = 1 (17)/(16) e² = 1 e² = (16)/(17) e = 4√(17)
Pattern Recognition

Standard Ellipse relations: For standard ellipses, the ratio of axes and the eccentricity are coupled quadratic equations. Expressing b/a as a function of e allows direct solving of the eccentricity.

Chapter Mix

Class 11 Mathematics: Conic Sections

Q67 jee_main_2025_02_april_evening Parabola
Let the point P of the focal chord PQ of the parabola y² = 16x be (1, -4). If the focus of the parabola divides the chord PQ in the ratio m : n, (m, n) = 1, then m² + n² is equal to:
  • A. 17
  • B. 10
  • C. 37
  • D. 26

Solution

Related Formula
Parametric coordinates on y² = 4ax: (at², 2at) Focal Chord relation: t₁ t₂ = -1 Section Formula: (xc, yc) = ( (m x₂ + n x₁)/(m+n), (m y₂ + n y₁)/(m+n) )
Core Logic

We find the parametric parameters of coordinates P and Q, obtain their Cartesian values, and then apply the section formula with the focus S to calculate the splitting ratio.

Step 1: Find coordinates of P and Q

For parabola y² = 16x, the focal parameter is a = 4. Focus is S(4, 0). Let P be (a t₁², 2a t₁) = (1, -4):

2a t₁ = -4 2(4) t₁ = -4 t₁ = -(1)/(2)

Since PQ is a focal chord, the parametric points are coupled:

t₁ t₂ = -1 t₂ = 2

Now, calculate the coordinates of Q:

Q ≡ (a t₂², 2 a t₂) = (4(4), 2(4)(2)) = (16, 16)
Step 2: Solve for the dividing ratio

Let the focus S(4, 0) divide the line segment PQ internally in the ratio λ : 1. Using the y-coordinate of the section formula:

yₛ = (λ yq + 1 yₚ)/(λ + 1) 0 = (λ(16) + 1(-4))/(λ + 1) 16λ - 4 = 0 λ = (1)/(4)

Thus, the focus S divides the chord internally in the ratio 1:4. Since (1, 4) = 1, we have m = 1 and n = 4:

m² + n² = 1² + 4² = 1 + 16 = 17
Pattern Recognition

Harmonic Mean Shortcut: In any parabola, the focus divides a focal chord internally into segments of lengths SP and SQ such that the semi-latus rectum 2a is the harmonic mean of these segments: (1)/(SP) + (1)/(SQ) = (1)/(a).

Chapter Mix

Class 11 Mathematics: Conic Sections

Q jee_main_2025_02_april_morning Properties of Hyperbola
Let one focus of the hyperbola H: (x²)/(a²) - (y²)/(b²) = 1 be at (√(10), 0) and the corresponding directrix be x = 9√(10). If e and l respectively are the eccentricity and the length of the latus rectum of H, then 9(e² + l) is equal to:
  • A. 14
  • B. 15
  • C. 16
  • D. 12

Solution

Related Formula

For a standard hyperbola: Focus: (± ae, 0) Directrix: x = ± (a)/(e) Eccentricity relation: (ae)² = a² + b² Length of latus rectum: l = (2b²)/(a)

Core Logic

Given ae = √(10) and (a)/(e) = 9√(10). Multiplying these gives a², which determines both parameters.

Step 1: Find a and e
a² = (ae) · ((a)/(e)) = √(10) · 9√(10) = 9 a = 3

Substitute a = 3 into ae = √(10):

e = √(10)3 e² = (10)/(9)
Step 2: Find b and l

Using (ae)² = a² + b²:

10 = 9 + b² b² = 1

Then the length of latus rectum l is:

l = (2b²)/(a) = (2(1))/(3) = (2)/(3)
Step 3: Evaluate Final Expression

Calculate 9(e² + l):

9((10)/(9) + (2)/(3)) = 10 + 6 = 16
Pattern Recognition

Multiplying focus location by directrix location immediately eliminates e, giving a² directly. Once a² is known, b² follow seamlessly via (ae)² = a²+b².

Chapter Mix

Class 11 Mathematics: Conic Sections

More Conic Sections Questions — jee_main_2025_29_jan_evening

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