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

2022 · 27 Jul · Shift 1 · Q41
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. /2022 · 27 Jul · Shift 1 · Q41

Hyperbola question

2022 · 27 Jul · Shift 1 · Q41

JEE MainMathematicsHyperbolaNumerical+4 / −1
An ellipse E:x2a2+y2b2=1E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1E:a2x2​+b2y2​=1 passes through the vertices of the hyperbola H:x249−y264=−1H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1H:49x2​−64y2​=−1. Let the major and minor axes of the ellipse EEE coincide with the transverse and conjugate axes of the hyperbola HHH, respectively. Let the product of the eccentricities of EEE and HHH be 12\frac{1}{2}21​. If lll is the length of the latus rectum of the ellipse EEE, then the value of 113l113 l113l is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1552

  1. Write the hyperbola in standard form

Given

x249−y264=−1\frac{x^2}{49}-\frac{y^2}{64}=-149x2​−64y2​=−1

Rewriting,

y264−x249=1\frac{y^2}{64}-\frac{x^2}{49}=164y2​−49x2​=1

So this is a hyperbola with:

  • transverse axis along the yyy-axis,
  • conjugate axis along the xxx-axis,
  • ah2=64,  bh2=49a_h^2=64,\; b_h^2=49ah2​=64,bh2​=49.

Hence its vertices are

(0,±8).(0,\pm 8).(0,±8).
  1. Use the condition about the ellipse axes

The major and minor axes of ellipse EEE coincide with the transverse and conjugate axes of the hyperbola respectively.

  • Hyperbola transverse axis is along yyy-axis.
  • Hyperbola conjugate axis is along xxx-axis.

Therefore ellipse EEE has:

  • major axis along yyy-axis,
  • minor axis along xxx-axis.

So the ellipse should be written as

x2b2+y2a2=1,a>b.\frac{x^2}{b^2}+\frac{y^2}{a^2}=1, \qquad a>b.b2x2​+a2y2​=1,a>b.

But the problem gives it as

x2a2+y2b2=1.\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.a2x2​+b2y2​=1.

Thus, for consistency with the axis condition, we must have the larger denominator under y2y^2y2. Since the ellipse passes through (0,±8)(0,\pm 8)(0,±8),

0a2+64b2=1impliesb2=64.\frac{0}{a^2}+\frac{64}{b^2}=1 implies b^2=64.a20​+b264​=1impliesb2=64.

So the semi-axis along yyy is 888, meaning actually the intended major semi-axis is 888.

Hence we interpret the ellipse as having semi-major axis 888 and semi-minor axis equal to the other parameter. To avoid notation confusion, let ellipse semi-major axis A=8A=8A=8 and semi-minor axis BBB.

Then

eE=1−B2A2=1−B264.e_E=\sqrt{1-\frac{B^2}{A^2}}=\sqrt{1-\frac{B^2}{64}}.eE​=1−A2B2​​=1−64B2​​.
  1. Eccentricity of the hyperbola

For hyperbola

y264−x249=1,\frac{y^2}{64}-\frac{x^2}{49}=1,64y2​−49x2​=1,

its eccentricity is

eH=1+bh2ah2=1+4964=11364=1138.e_H=\sqrt{1+\frac{b_h^2}{a_h^2}}=\sqrt{1+\frac{49}{64}}=\sqrt{\frac{113}{64}}=\frac{\sqrt{113}}{8}.eH​=1+ah2​bh2​​​=1+6449​​=64113​​=8113​​.
  1. Use the product of eccentricities

Given

eEeH=12.e_E e_H=\frac12.eE​eH​=21​.

So,

eE⋅1138=12e_E\cdot \frac{\sqrt{113}}{8}=\frac12eE​⋅8113​​=21​

which gives

eE=4113.e_E=\frac{4}{\sqrt{113}}.eE​=113​4​.

Now square both sides:

1−B264=16113.1-\frac{B^2}{64}=\frac{16}{113}.1−64B2​=11316​.

Therefore

B264=1−16113=97113.\frac{B^2}{64}=1-\frac{16}{113}=\frac{97}{113}.64B2​=1−11316​=11397​.

So

B2=64⋅97113=6208113.B^2=64\cdot \frac{97}{113}=\frac{6208}{113}.B2=64⋅11397​=1136208​.
  1. Find the latus rectum of the ellipse

For an ellipse with semi-major axis AAA and semi-minor axis BBB, length of latus rectum is

l=2B2A.l=\frac{2B^2}{A}.l=A2B2​.

Here A=8A=8A=8, so

l=2(6208113)8=6208452=1552113.l=\frac{2\left(\frac{6208}{113}\right)}{8}=\frac{6208}{452}=\frac{1552}{113}.l=82(1136208​)​=4526208​=1131552​.

Hence

113l=1552.113l=1552.113l=1552.
  1. Comparison with stored answer

Derived answer:

113l=1552.113l=1552.113l=1552.

This matches the stored correct answer.

PreviousNext

More from Hyperbola

  • 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
  • The locus of the point of intersection of the lines (3​)kx+ky−43​=0 and 3​x−y−4(3​)k=0 is a conic, whose eccentricity is ​.2021 · Numerical
  • A hyperbola passes through the foci of the ellipse 25x2​+16y2​=1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their…2021 · MCQ
  • The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2−9y2+32x+36y−164=0, and its foci is :2021 · MCQ
  • The locus of the mid points of the chords of the hyperbola x2 − y2 = 4, which touch the parabola y2 = 8x, is :2021 · MCQ