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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Hyperbola

NDA 2016 · Mathematics · Hyperbola

What is the equation of the hyperbola having latus rectum and...

What is the equation of the hyperbola having latus rectum and eccentrieity 8 and \frac{3}{\sqrt{5}} respectively?

Options

  1. Ax225−y220=1\frac{x^2}{25}-\frac{y^2}{20}=1
  2. Bx240−y220=1\frac{x^2}{40}-\frac{y^2}{20}=1
  3. Cx240−y230=1\frac{x^2}{40}-\frac{y^2}{30}=1
  4. Dx230−y225=1\frac{x^2}{30}-\frac{y^2}{25}=1

Correct answer

A. x225−y220=1\frac{x^2}{25}-\frac{y^2}{20}=1

Step-by-step solution

Let the equation of hyperbola be x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 Latus rectum =8=2b2a⇒b2=4a...(i)=8=\frac{2 b^2}{a} \Rightarrow \mathrm{ b}^2=4 \mathrm{a} ...(i) Also, b2=a2(e2−1)⇒4a=a2(e2−1)⇒4a=a2[(35)2−1]⇒a=5b2=20∴ Equation is x225−y220=1\begin{aligned} & \text{ Also, } \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right) \\ \Rightarrow & 4 \mathrm{a}=\mathrm{a}^2\left(\mathrm{e}^2-1\right) \\ \Rightarrow & 4 \mathrm{a}=\mathrm{a}^2\left[\left(\frac{3}{\sqrt{5}}\right)^2-1\right] \\ \Rightarrow & \mathrm{a}=5 & \mathrm{ b}^2=20 \\ \therefore & \text{ Equation is } \frac{x^2}{25}-\frac{y^2}{20}=1 \end{aligned}$$ [From(i)]
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