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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Hyperbola

NDA 2015 · Mathematics · Hyperbola

The hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 passes through the point (3...

The hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 passes through the point (35,1)(3 \sqrt{5}, 1) and the length of its latus rectum is 43\frac{4}{3} units. The length of the conjugate axis is

Options

  1. A2 units
  2. B3 units
  3. C4 units
  4. D5 units

Correct answer

C. 4 units

Step-by-step solution

x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 Hyperbola passes through (35,1)(3 \sqrt{5}, 1)∴(35)2a2−1b2=145a2−1b2=1...(i)\begin{aligned} \therefore \quad & \frac{(3 \sqrt{5})^2}{\mathrm{a}^2}-\frac{1}{\mathrm{ b}^2}=1 \\ & \frac{45}{\mathrm{a}^2}-\frac{1}{\mathrm{ b}^2}=1...(i) \end{aligned}$\begin{aligned} & \text{ Now length\text{ of } latus rectum }=\frac{2 b^2}{a} \\ \Rightarrow & \frac{4}{3}=\frac{2 b^2}{\mathrm{a}} \\ \Rightarrow & \frac{2}{3}=\frac{\mathrm{b}^2}{\mathrm{a}} \Rightarrow \mathrm{a}=\frac{3 \mathrm{ b}^2}{2}...(ii) \end{aligned} Putting the value of ‘a’ from equation (ii) in equation (i), ⇒45×49b4−1b2=1⇒20b4−1b2=1\begin{aligned} & \Rightarrow \quad \frac{45 \times 4}{9 b^4}-\frac{1}{b^2}=1 \\ & \Rightarrow \quad \frac{20}{b^4}-\frac{1}{b^2}=1\end{aligned}20−b2=b4b4+b2−20=0b4+5b2−4b2−20=0b2(b2+5)−4(b2+5)=0(b2−4)(b2+5)=0b2=4,b2=−5∴b2=4⇒b=2\begin{aligned} & 20-b^2=b^4 \\ & b^4+b^2-20=0 \\ & b^4+5 b^2-4 b^2-20=0 \\ & b^2\left(b^2+5\right)-4\left(b^2+5\right)=0 \\ & \left(b^2-4\right)\left(b^2+5\right)=0 \\ & b^2=4, b^2=-5 \\ & \therefore \quad b^2=4 \Rightarrow b=2 \end{aligned}NowlengthofconjugateaxisNow length of conjugate axis=2 b=2(2)=4\therefore \quad$ Option (C) is correct.
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