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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

If e_1 is the eccentricity of the ellipse (x^2)/(16)+(y^2)/(25)=1...

If e1e_1 is the eccentricity of the ellipse x216+y225=1\frac{x^2}{16}+\frac{y^2}{25}=1 and e2e_2 is the eccentricity of a hyperbola passing through the foci of the given ellipse and e1e2=1e_1 e_2=1, then the equation of such a hyperbola among the following is

Options

  1. Ax29−y216=1\frac{x^2}{9}-\frac{y^2}{16}=1
  2. By29−x216=1\frac{y^2}{9}-\frac{x^2}{16}=1
  3. Cx29−y225=1\frac{x^2}{9}-\frac{y^2}{25}=1
  4. Dx225−y29=1\frac{x^2}{25}-\frac{y^2}{9}=1

Correct answer

B. y29−x216=1\frac{y^2}{9}-\frac{x^2}{16}=1

Step-by-step solution

Equation of ellipse x216+y225=1\frac{x^2}{16}+\frac{y^2}{25}=1 Foci =(0,±3)⇒e2=1−1625=35\quad=(0, \pm 3)\Rightarrow \quad e_2=\sqrt{1-\frac{16}{25}}=\frac{3}{5} given e2e2=1e_2 e_2=1 (where e2e_2 is eccentricity of hyperbola) Let equation of hyperbola −x2a2+y2b2=1-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 its passes through (0,±3)(0, \pm 3)∴b2=9⇒e2=1+a2b2⇒e22=1+a2b2⇒1e22=1+a29⇒259=1+a29⇒a2=16\begin{array}{ll} \therefore& b²=9 \\ \Rightarrow & e_2=\sqrt{1+\frac{a^2}{b^2}}\\ \Rightarrow & e_2^2=1+\frac{a^2}{b^2}\\ \Rightarrow & \frac{1}{e_2^2}=1+\frac{a^2}{9} \Rightarrow \frac{25}{9}=1+\frac{a^2}{9}\\ \Rightarrow & a²=16 \end{array}Hence,theequationofhyperbolaisHence, the equation of hyperbola is\frac{y^2}{9}-\frac{x^2}{16}=1
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