Conic Sections: JEE Main PYQ, 2020–2025

Answer: (A)

Compare y² = 12x with y² = 4ax to get a = 3. The parametric point is (at², 2at). For t = 2 this is (3 × 4, 2 × 3 × 2) = (12, 12). Check: 12² = 144 = 12 × 12.

Answer: (A)

The circles have centres C₁(0, 0) with radius 1 and C₂(10, 0) with radius 5. For external contact with the variable circle, PC₁ = r + 1 and PC₂ = r + 5, so PC₂ − PC₁ = 4, a constant smaller than C₁C₂ = 10. By the definition of a hyperbola, P lies on a hyperbola with foci C₁ and C₂, with 2a = 4 and 2c = 10. So a = 2, c = 5 and b² = c² − a² = 21, and the centre is the midpoint (5, 0). The equation is (x − 5)²/4 − y²/21 = 1, and P lies on the branch nearer C₁.

Answer: (A)

The perimeter is 18 and AB = 8, so CA + CB = 10, a constant greater than AB. So C lies on an ellipse with foci A and B, with 2a = 10 and 2c = 8. Then a = 5, c = 4 and b² = 9, giving x²/25 + y²/9 = 1 (excluding the points on the x-axis). The centroid is G = ((−4 + 4 + x)/3, y/3) = (x/3, y/3). Substituting x = 3X and y = 3Y gives X²/(25/9) + Y² = 1. Here a² = 25/9 and b² = 1, so c² = 16/9 and c = 4/3. The distance between the foci is 2c = 8/3.

Answer: (A)

For y² = 8x, 4a = 8, so a = 2 and the focus is (2, 0). The latus rectum is the line x = 2, which meets the parabola where y² = 16, so y = ±4. The triangle has vertices (0, 0), (2, 4) and (2, −4). Its base is 8 and its height is 2, so the area is ½ × 8 × 2 = 8.

Answer: (A)

With the axis along the y-axis and the curve passing through a point below the x-axis, the form is x² = −4ay. Substitute (4, −2): 16 = −4a(−2) = 8a, so a = 2. The parabola is x² = −8y, the focus is (0, −2), and the directrix is y = a = 2.

Answer: (A)

For y² = 12x, a = 3 and the focus is (3, 0). The chord through the focus with slope 2 is x = 3 + y/2. Substituting into y² = 12x gives y² = 36 + 6y, so y² − 6y − 36 = 0. The roots satisfy y₁ + y₂ = 6 and y₁y₂ = −36, so the midpoint has y = 3 and x = 3 + 3/2 = 9/2. The centre is (9/2, 3). The chord length is √(1 + 1/4)·|y₁ − y₂| = (√5/2)·√(36 + 144) = (√5/2)(6√5) = 15. This agrees with the focal chord formula 4a/sin²θ = 12/(4/5) = 15, since tanθ = 2 gives sin²θ = 4/5. The radius is 15/2, so the circle is (x − 9/2)² + (y − 3)² = 225/4. As a check, the distance from the centre to the directrix x = −3 is 9/2 + 3 = 15/2, which equals the radius.

Answer: (A)

The larger denominator gives a², so a² = 25 and b² = 16. Then c² = a² − b² = 9 and c = 3, so e = c/a = 3/5. The major axis is along the y-axis here, which does not change the formula.

Answer: (A)

For the ellipse, a_e = 13 and b_e = 12, so c_e² = 169 − 144 = 25 and the foci are (±5, 0). The vertices are (±13, 0). The hyperbola's vertices are therefore (±5, 0), so a = 5, and its foci are (±13, 0), so c = 13. For a hyperbola b² = c² − a² = 169 − 25 = 144, so b = 12 and the conjugate axis has length 2b = 24.

Answer: (A)

For any point P on a hyperbola, |PF₁ − PF₂| = 2a. Here a = 3, b = 4, so c = √(9 + 16) = 5 and F₁F₂ = 10. Since angle F₁PF₂ = 90°, the Pythagorean theorem gives PF₁² + PF₂² = 100. Squaring the difference of focal distances gives 36 = PF₁² + PF₂² − 2·PF₁·PF₂ = 100 − 2·PF₁·PF₂, so PF₁·PF₂ = 32. The area is ½·PF₁·PF₂ = 16. Such a P exists because the circle x² + y² = 25 meets the hyperbola at x² = 369/25, which is greater than 9. In general, the area is b²cot(θ/2) for angle θ at P, and θ = 90° gives b².

Answer: (A)

For x²/a² − y²/b² = 1, the vertices are (±a, 0) and the foci are (±c, 0) with c² = a² + b². Here a = 3 and c = 5, so b² = 25 − 9 = 16. The equation is x²/9 − y²/16 = 1.