Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Hyperbola question

2023 · 31 Jan · Shift 2 · Q34
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Hyperbola
  5. /2023 · 31 Jan · Shift 2 · Q34

Hyperbola question

2023 · 31 Jan · Shift 2 · Q34

JEE MainMathematicsHyperbolaMCQ+4 / −1
Let H\mathrm{H}H be the hyperbola, whose foci are (1±2,0)(1 \pm \sqrt{2}, 0)(1±2​,0) and eccentricity is 2\sqrt{2}2​. Then the length of its latus rectum is :
  1. A
    52\frac{5}{2}25​
  2. B
    3
  3. C
    2
  4. D
    32\frac{3}{2}23​
View written solutionFree

Correct answer: C

  1. Identify the center and transverse axis

The foci are given as

(1±2,0).(1\pm \sqrt{2},0).(1±2​,0).

So the center is the midpoint of the foci:

((1+2)+(1−2)2,0+02)=(1,0).\left(\frac{(1+\sqrt2)+(1-\sqrt2)}2,\frac{0+0}2\right)=(1,0).(2(1+2​)+(1−2​)​,20+0​)=(1,0).

Since both foci lie on the horizontal line y=0y=0y=0, the hyperbola has horizontal transverse axis.

Thus its standard form is

(x−1)2a2−y2b2=1.\frac{(x-1)^2}{a^2}-\frac{y^2}{b^2}=1.a2(x−1)2​−b2y2​=1.
  1. Use the focal distance

For a hyperbola,

c2=a2+b2,c^2=a^2+b^2,c2=a2+b2,

and eccentricity is

e=ca.e=\frac{c}{a}.e=ac​.

From the foci, the distance of each focus from the center is

c=2.c=\sqrt2.c=2​.

Given eccentricity

e=2.e=\sqrt2.e=2​.

So

ca=2.\frac{c}{a}=\sqrt2.ac​=2​.

Substitute c=2c=\sqrt2c=2​:

2a=2  ⟹  a=1.\frac{\sqrt2}{a}=\sqrt2 \implies a=1.a2​​=2​⟹a=1.

Hence,

a2=1.a^2=1.a2=1.
  1. Find b2b^2b2

Using

c2=a2+b2,c^2=a^2+b^2,c2=a2+b2,

we get

2=1+b2  ⟹  b2=1.2=1+b^2 \implies b^2=1.2=1+b2⟹b2=1.
  1. Length of latus rectum

For the hyperbola

(x−h)2a2−(y−k)2b2=1,\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,a2(x−h)2​−b2(y−k)2​=1,

the length of the latus rectum is

2b2a.\frac{2b^2}{a}.a2b2​.

Substitute a=1a=1a=1 and b2=1b^2=1b2=1:

Length of latus rectum=2⋅11=2.\text{Length of latus rectum}=\frac{2\cdot 1}{1}=2.Length of latus rectum=12⋅1​=2.
  1. Match with options
222

corresponds to Option C.

PreviousNext

More from Hyperbola

  • Let the hyperbola H:a2x2​−y2=1 and the ellipse E:3x2+4y2=12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH​ and eE​ are the eccentricities of H…2022 · Numerical
  • Let the foci of the ellipse 16x2​+7y2​=1 and the hyperbola 144x2​−αy2​=251​ coincide. Then the length of the latus rectum of the hyperbola is :2022 · MCQ
  • If the line x−1=0 is a directrix of the hyperbola kx2−y2=6, then the hyperbola passes through the point :2022 · MCQ
  • An ellipse E:a2x2​+b2y2​=1 passes through the vertices of the hyperbola H:49x2​−64y2​=−1. Let the major and minor axes of the ellipse E coincide with the transverse and conjugate…2022 · Numerical
  • Let the hyperbola H:a2x2​−b2y2​=1 pass through the point (22​,−22​). A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the…2022 · MCQ
  • Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola a2x2​−b2y2​=1. Let e' and l' respectively be the eccentricity and length of…2022 · MCQ
  • Let H:a2x2​−b2y2​=1, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(22​+14​). If the eccentricity H is 211​​…2022 · Numerical
  • The locus of the midpoints of the chord of the circle, x2 + y2 = 25 which is tangent to the hyperbola, 9x2​−16y2​=1 is :2021 · MCQ