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x – 2y + 4 = 0 is a common tangent to y2 = 4x & 
+ 
= 1. Then the value of b and the other common tangent are given by
+ = 1. Then the value of b and the other common tangent are given by- a)b = √3 ;x + 2y + 4 = 0
- b)b = 3; x + 2y + 4 = 0
- c)b = √3 ;x + 2y– 4 = 0
- d)b = √3 ; x – 2y – 4 = 0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the ...
Equation of tangent of ellipse is
y=mx±+ ..............(i)
Given equation is x−2y+4=0 .........(ii)
Since (i) & (ii) are same, comparing them, we get
m=1/2&a2m2 + b2 = 2
⇒4*1/4+b2=4
b=±3
Equation of tangent of parabola
y=mx+1/m .........(iii)
by (i) & (iii)
1/m^2= a2m2+b2
on solving it we get m=±1/2
with m = −1/2 , we get x+2y+4=0
y=mx±+ ..............(i)
Given equation is x−2y+4=0 .........(ii)
Since (i) & (ii) are same, comparing them, we get
m=1/2&a2m2 + b2 = 2
⇒4*1/4+b2=4
b=±3
Equation of tangent of parabola
y=mx+1/m .........(iii)
by (i) & (iii)
1/m^2= a2m2+b2
on solving it we get m=±1/2
with m = −1/2 , we get x+2y+4=0
Most Upvoted Answer
x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the ...
Equation of tangent of ellipse is
y=mx±+ ..............(i)
Given equation is x−2y+4=0 .........(ii)
Since (i) & (ii) are same, comparing them, we get
m=1/2&a2m2 + b2 = 2
⇒4*1/4+b2=4
b=±3
Equation of tangent of parabola
y=mx+1/m .........(iii)
by (i) & (iii)
1/m^2= a2m2+b2
on solving it we get m=±1/2
with m = −1/2 , we get x+2y+4=0
y=mx±+ ..............(i)
Given equation is x−2y+4=0 .........(ii)
Since (i) & (ii) are same, comparing them, we get
m=1/2&a2m2 + b2 = 2
⇒4*1/4+b2=4
b=±3
Equation of tangent of parabola
y=mx+1/m .........(iii)
by (i) & (iii)
1/m^2= a2m2+b2
on solving it we get m=±1/2
with m = −1/2 , we get x+2y+4=0
Community Answer
x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the ...
Use Cond Of tangency To find B Then Apply This In Other parts to get right tangent
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Question Description
x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2026 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer?.
x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2026 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer?.
Solutions for x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 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 x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice x –2y + 4 = 0 is a common tangent to y2= 4x &+= 1. Then the value of b and the other common tangent are given bya)b = √3 ;x + 2y + 4 = 0b)b = 3; x + 2y + 4 = 0c)b = √3;x + 2y–4 = 0d)b = √3; x –2y –4 = 0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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