Let L₁ be a tangent to the parabola y²=4(x+1) and L₂ be a tangent to the parabola y²=8(x+2) such that L₁ and L₂ intersect at right angles. Then L₁ and L₂ meet on the straight line:
Let L₁ be a tangent to the parabola y²=4(x+1) and L₂ be a tangent to the parabola y²=8(x+2) such that L₁ and L₂ intersect at right angles. Then L₁ and L₂ meet on the straight line:
Option 1 -
x+3=0
Option 2 -
2x+1=0
Option 3 -
x+2=0
Option 4 -
x+2y=0
-
1 Answer
-
Correct Option - 1
Detailed Solution:Equation of tangent to y²=4 (x+1) is y=m (x+1)+1/m.
Equation of tangent to y²=8 (x+2) is y=m' (x+2)+2/m'.
m'=-1/m.
Solving for intersection point, x+3=0.
Similar Questions for you
Eqn : y – 0 = tan45° (x – 9) Þ y = (x – 9)
Option (B) is correct
|r1 – r2| < c1c2 < r1 + r2
->
Now,
(y – 2) = m (x – 8)
⇒ x-intercept
⇒
⇒ y-intercept
⇒ (–8m + 2)
⇒ OA + OB =
->
->
->
->Minimum = 18
Kindly consider the following figure
According to question,
Equation of required line is
Obviously B (2, 2) satisfying condition (i)
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers