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

2020 · 2 Sep · Shift 2 · Q39
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. /2020 · 2 Sep · Shift 2 · Q39

Hyperbola question

2020 · 2 Sep · Shift 2 · Q39

JEE MainMathematicsHyperbolaMCQ+4 / −1
For some θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right)θ∈(0,2π​), if the eccentricity of the hyperbola, x2–y2sec2 θ\thetaθ= 10 is 5\sqrt 55​ times the eccentricity of the ellipse, x2sec2 θ\thetaθ + y2 = 5, then the length of the latus rectum of the ellipse, is :
  1. A
    30\sqrt {30}30​
  2. B
    262\sqrt 626​
  3. C
    453{{4\sqrt 5 } \over 3}345​​
  4. D
    253{{2\sqrt 5 } \over 3}325​​
View written solutionFree

Correct answer: C

  1. Write both conics in standard form

Given hyperbola: x2−y2sec⁡2θ=10x^2-y^2\sec^2\theta=10x2−y2sec2θ=10 Divide by 101010:

\implies \frac{x^2}{10}-\frac{y^2}{10\cos^2\theta}=1$$ So for the hyperbola, $$a_h^2=10,\qquad b_h^2=10\cos^2\theta$$ Hence its eccentricity is $$e_h=\sqrt{1+\frac{b_h^2}{a_h^2}}=\sqrt{1+\cos^2\theta}$$ --- Given ellipse: $$x^2\sec^2\theta+y^2=5$$ Divide by $5$: $$\frac{x^2\sec^2\theta}{5}+\frac{y^2}{5}=1 \implies \frac{x^2}{5\cos^2\theta}+\frac{y^2}{5}=1$$ Since $\theta\in(0,\pi/2)$, we have $\cos^2\theta<1$, so major axis is along $y$. Thus, $$a_e^2=5,\qquad b_e^2=5\cos^2\theta$$ Its eccentricity is $$e_e=\sqrt{1-\frac{b_e^2}{a_e^2}}=\sqrt{1-\cos^2\theta}=\sin\theta$$ --- 2. **Use the given relation between eccentricities** Given: $$e_h=\sqrt{5}\,e_e$$ So, $$\sqrt{1+\cos^2\theta}=\sqrt{5}\sin\theta$$ Squaring both sides, $$1+\cos^2\theta=5\sin^2\theta$$ Using $\sin^2\theta=1-\cos^2\theta$, $$1+\cos^2\theta=5(1-\cos^2\theta)$$ $$1+\cos^2\theta=5-5\cos^2\theta$$ $$6\cos^2\theta=4$$ $$\cos^2\theta=\frac{2}{3}$$ Then, $$\sin^2\theta=1-\frac{2}{3}=\frac{1}{3}$$ --- 3. **Find the latus rectum of the ellipse** For an ellipse, length of latus rectum is $$\ell=\frac{2b^2}{a}$$ where $a$ is the semi-major axis and $b$ is the semi-minor axis. Here, $$a=\sqrt{5},\qquad b^2=5\cos^2\theta=5\cdot\frac{2}{3}=\frac{10}{3}$$ Therefore, $$\ell=\frac{2\cdot \frac{10}{3}}{\sqrt{5}}=\frac{20}{3\sqrt{5}}=\frac{4\sqrt{5}}{3}$$ --- 4. **Match with options** $$\boxed{\frac{4\sqrt{5}}{3}}$$ So the correct option is **C**.
PreviousNext

More from Hyperbola

  • A hyperbola having the transverse axis of length 2​ has the same foci as that of the ellipse 3x2 + 4y2 = 12, then this hyperbola does not pass through which of the following points?2020 · MCQ
  • Let e1 and e2 be the eccentricities of the ellipse, 25x2​+b2y2​=1(b < 5) and the hyperbola, 16x2​−b2y2​=1 respectively satisfying e1e2 = 1. If α and β…2020 · MCQ
  • If e1 and e2 are the eccentricities of the ellipse, 18x2​+4y2​=1 and the hyperbola, 9x2​−4y2​=1 respectively and (e1, e2) is a point on the ellipse, 15x2 + 3y2 = k, then k…2020 · MCQ
  • Let 0<θ<2π​. If the eccentricity of the hyperbola cos2θx2​−sin2θy2​ = 1 is greater than 2, then the length of its latus rectum lies in the interval :2019 · MCQ
  • A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :2019 · MCQ
  • If a directrix of a hyperbola centred at the origin and passing through the point (4, –2 3​ ) is 5x = 4 5​ and its eccentricity is e, then :2019 · MCQ
  • If 5x + 9 = 0 is the directrix of the hyperbola 16x2 – 9y2 = 144, then its corresponding focus is :2019 · MCQ
  • If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is :2019 · MCQ