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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced · Mathematics · Hyperbola

If a hyperbola whose foci are (-2,4) and (4,6) touches y -axis then...

If a hyperbola whose foci are (−2,4)(-2,4) and (4,6)(4,6) touches yy-axis then equation of hyperbola is

Options

  1. A(3x+y−8)2−(x−3y+14)24=20(3 x+y-8)^2-\frac{(x-3 y+14)^2}{4}=20
  2. B(x+3y−7)22−(x−3y+8)28=1\frac{(x+3 y-7)^2}{2}-\frac{(x-3 y+8)^2}{8}=1
  3. C(3x+y−8)22−(x−3y+14)28=1\frac{(3 x+y-8)^2}{2}-\frac{(x-3 y+14)^2}{8}=1
  4. D(x−3y+14)22−(3x+y−8)28=20\frac{(x-3 y+14)^2}{2}-\frac{(3 x+y-8)^2}{8}=20

Correct answer

A. (3x+y−8)2−(x−3y+14)24=20(3 x+y-8)^2-\frac{(x-3 y+14)^2}{4}=20

Step-by-step solution

b2=4×2=8,2ae=40a2e2=10=b2+a2\begin{aligned} & b^2=4 \times 2=8,2 a e=\sqrt{40} \\ & a^2 e^2=10=b^2+a^2\end{aligned}⇒a2=2 Centre =(1,5) Equation of transverse axis ≡x−3y+14=0\begin{aligned} & \Rightarrow a^2=2 \\ & \text { Centre }=(1,5) \\ & \text { Equation of transverse axis } \equiv x-3 y+14=0\end{aligned}EquationofconjugateaxisEquation of conjugate axis\equiv 3 x+y-8=0EquationofhyperbolaEquation of hyperbola=\frac{(3 x+y-8)^2}{1}-\frac{(x-3 y+14)^2}{4}=20$
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