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The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 is
- a)
- 
= 1 - b)
- 
= 1 - c)
- 
= - 1 - d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The equation of the hyperbola whose foci are (6, 5), (–4, 5) and...
Let the centre of hyperbola be (α,β)
As y=5 line has the foci, it also has the major axis.
∴ [(x−α)2]/a2 − [(y−β)2]/b2 = 1
Midpoint of foci = centre of hyperbola
∴ α=1,β=5
Given, e= 5/4
We know that foci is given by (α±ae,β)
∴ α+ae=6
⇒1+(5/4a)=6
⇒ a=4
Using b2 = a2(e2 − 1)
⇒ b2=16((25/16)−1)=9
∴ Equation of hyperbola ⇒ [(x−1)2]/16−[(y−5)2]/9=1
As y=5 line has the foci, it also has the major axis.
∴ [(x−α)2]/a2 − [(y−β)2]/b2 = 1
Midpoint of foci = centre of hyperbola
∴ α=1,β=5
Given, e= 5/4
We know that foci is given by (α±ae,β)
∴ α+ae=6
⇒1+(5/4a)=6
⇒ a=4
Using b2 = a2(e2 − 1)
⇒ b2=16((25/16)−1)=9
∴ Equation of hyperbola ⇒ [(x−1)2]/16−[(y−5)2]/9=1
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The equation of the hyperbola whose foci are (6, 5), (–4, 5) and...
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The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 isa)-= 1b)-= 1c)-= - 1d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 isa)-= 1b)-= 1c)-= - 1d)None of theseCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 isa)-= 1b)-= 1c)-= - 1d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 isa)-= 1b)-= 1c)-= - 1d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
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