JEE Exam > JEE Questions > The equation of the common tangent to the par...
Start Learning for Free
The equation of the common tangent to the parabola y2 = 8x and the hyperbola 3x2– y2 = 3 is
- a)2x ± y + 1 = 0
- b)x ± y + 1 = 0
- c)x ± 2y + 1 = 0
- d)x ± y + 2 = 0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The equation of the common tangent to the parabola y2= 8x and the hype...
Most Upvoted Answer
The equation of the common tangent to the parabola y2= 8x and the hype...
Explanation:
Equation of Parabola:
- The given equation of the parabola is y^2 = 8x.
- This is a parabola that opens towards the right.
Equation of Hyperbola:
- The given equation of the hyperbola is 3x^2 - y^2 = 3.
- This is a hyperbola that opens towards the x-axis.
Finding the Tangent:
- To find the common tangent to both curves, we need to find the point of tangency.
- Differentiating the equations of the curves, we get:
- For the parabola: dy/dx = 4/y
- For the hyperbola: dy/dx = 2x/2y = x/y
- Equating the derivatives and solving for x and y, we get x = 1, y = ±2.
Equation of Tangent:
- The equation of the tangent to a curve at a point (x1, y1) is given by:
- (y - y1) = m(x - x1), where m is the slope of the tangent.
- Substituting the point of tangency (1, 2) into the equations of the parabola and hyperbola, we get the slope of the tangent as 1/2.
- Therefore, the equation of the common tangent is:
- y - 2 = 1/2(x - 1)
- 2x - y - 3 = 0
- 2x - y + 1 = 0
Therefore, the correct answer is option 'A': 2x ± y + 1 = 0.
Equation of Parabola:
- The given equation of the parabola is y^2 = 8x.
- This is a parabola that opens towards the right.
Equation of Hyperbola:
- The given equation of the hyperbola is 3x^2 - y^2 = 3.
- This is a hyperbola that opens towards the x-axis.
Finding the Tangent:
- To find the common tangent to both curves, we need to find the point of tangency.
- Differentiating the equations of the curves, we get:
- For the parabola: dy/dx = 4/y
- For the hyperbola: dy/dx = 2x/2y = x/y
- Equating the derivatives and solving for x and y, we get x = 1, y = ±2.
Equation of Tangent:
- The equation of the tangent to a curve at a point (x1, y1) is given by:
- (y - y1) = m(x - x1), where m is the slope of the tangent.
- Substituting the point of tangency (1, 2) into the equations of the parabola and hyperbola, we get the slope of the tangent as 1/2.
- Therefore, the equation of the common tangent is:
- y - 2 = 1/2(x - 1)
- 2x - y - 3 = 0
- 2x - y + 1 = 0
Therefore, the correct answer is option 'A': 2x ± y + 1 = 0.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer?
Question Description
The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer?.
The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer?.
Solutions for The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The equation of the common tangent to the parabola y2= 8x and the hyperbola 3x2–y2= 3 isa)2x ± y + 1 = 0b)x ± y + 1 = 0c)x ± 2y + 1 = 0d)x ± y + 2 = 0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup