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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

For the hyperbola x^2-y^2-4 x+2 y+c=0 , if the focus is S(2+2 2 ,...

For the hyperbola x2−y2−4x+2y+c=0x^2-y^2-4 x+2 y+c=0, if the focus is S(2+22,k)S(2+2 \sqrt{2}, k) and the directrix that is adjacent to SS is x=2+2x=2+\sqrt{2}, then c=c=

Options

  1. A0
  2. B-1
  3. C1
  4. D2

Correct answer

B. -1

Step-by-step solution

Given hyperbola x2−y2−4x+2y+c=0(x−2)2−(y−1)2=3−c(x−2)23−c−(y−1)23−c=1\begin{aligned} x^2-y^2-4 x+2 y+c & =0 \\ (x-2)^2-(y-1)^2 & =3-c \\ \frac{(x-2)^2}{3-c}-\frac{(y-1)^2}{3-c} & =1 \end{aligned} Focus of hyperbola is x−2=±(3−c)+(3−c)x−2=±6−2c[∵ae=(3−c)+(3−c)]x=2±6−2c\begin{aligned} x-2 & = \pm \sqrt{(3-c)+(3-c)} \\ x-2 & = \pm \sqrt{6-2 c} \quad[\because a e=\sqrt{(3-c)+(3-c)}] \\ x & =2 \pm \sqrt{6-2 c} \end{aligned} Here, 2+6−2c=2+22∴6−2c=22⇒6−2c=8⇒c=−12+\sqrt{6-2 c}=2+2 \sqrt{2}\therefore \quad \sqrt{6-2 c}=2 \sqrt{2} \Rightarrow 6-2 c=8 \Rightarrow c=-1
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