Q71.If a directrix of a hyperbola centered at the origin and passing through the point (4, −2√3) is and its eccentricity is e, then: (1) 4e4 + 8e2 −35 = 0 (2) 4e4 −24e2 + 35 = 0 (3) 4e4 −24e2 + 27 = 0 (4) 4e4 −12e2 −27 = 0 x4−1
What This Question Tests
This problem involves using the properties of a hyperbola including its directrix, eccentricity, and a point it passes through, to form equations and solve for the eccentricity.
Concepts Tested
Formulas Used
x²/a² - y²/b² = 1
Directrix x = a/e
b² = a²(e²-1)
📚 NCERT Sections This Tests
9.15 — Apply Mirror Equation And The Condition:
Physics Class 12 · Chapter 9
9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.
12.5 — A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,
Physics Class 12 · Chapter 12
12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
📋 Question Details
- Chapter
- Hyperbola
- Topic
- Properties of a hyperbola
- Year
- 2019
- Shift
- 10 Apr Shift 1
- Q Number
- Q71
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 11: Conic Sections
More from this Chapter
Q96.For the hyperbola = 1 , which of the following remains constant when α varies? cos2 α α − sin2 (1) eccentricity (2) directrix (3) abscissae of vertices (4) abscissae of foci
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Q71.A tangent to the hyperbola x2 meets x-axis at P and y-axis at Q. Lines PR and QR are drawn such 4 −y22 = 1 that OPRQ is a rectangle (where O is the origin). Then R lies on : (1) 4 + 2 = 1 (2) 2 − 4 = 1 x2 y2 x2 y2 (3) 2 + 4 = 1 (4) 4 − 2 = 1 x2 y2 x2 y2
Q72.Let P(3 sec θ, 2 tan θ) and Q(3 sec ϕ, 2 tan ϕ) where θ + ϕ = π2 , be two distinct points on the hyperbola x2 . Then the ordinate of the point of intersection of the normals at P and Q is: 9 −y24 = 1 (1) 11 3 (2) −113 (3) 13 2 (4) −132 = 5, then k is equal to: