Q66.A hyperbola passes through the foci of the ellipse x2 = 1 and its transverse and conjugate axes coincide 25 + 16 with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is: (1) x2 = 9 9 −y216 = 1 (2) x2 −y2 (3) x2 9 −y225 = 1 (4) x29 −y24 = 1
What This Question Tests
This question requires applying the definitions and properties of ellipses and hyperbolas, including their foci and eccentricities, to determine the equation of the hyperbola.
Concepts Tested
Formulas Used
For ellipse x^2/a^2 + y^2/b^2 = 1, foci = (±ae, 0)
e^2 = 1 - b^2/a^2
For hyperbola x^2/A^2 - y^2/B^2 = 1, foci = (±Ae_h, 0)
e_h^2 = 1 + B^2/A^2
📚 NCERT Sections This Tests
9.8 — A Beam Of Light Converges At A Point P. Now A Lens Is Placed In The
Physics Class 12 · Chapter 9
9.8 A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
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.
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.
📋 Question Details
- Chapter
- Hyperbola
- Topic
- Equation of hyperbola, eccentricity
- Year
- 2021
- Shift
- 25 Feb Shift 2
- Q Number
- Q66
- 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
Q71.If the eccentricity of a hyperbola x2 K 2 is = 1, which passes through (K, 2), is √133 , then the value of 9 −y2b2 (1) 18 (2) 8 (3) 1 (4) 2
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: