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​Q. IF THE LINE JOINING THE POINTS (0,3) AND (5,-2) IS A TANGENT TO THE CURVE- y=ax/1+x , THEN- a= 4/2(3) 1/2 a= 8/(3) 1/2 a= -1 +-( plus – minus) (3) 1/2 -2(plus – minus) 2(3) 1/2

​Q. IF THE LINE JOINING THE POINTS
(0,3) AND (5,-2) IS A TANGENT TO THE CURVE-
y=ax/1+x , THEN-
  1. a= 4/2(3)1/2
  2. a= 8/(3)1/2
  3. a= -1 +-( plus – minus) (3)1/2
  4. -2(plus – minus) 2(3)1/2
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

Slope of tangent:
m = \frac{-2-3}{5}
m = -1
y = \frac{ax}{1+x}
y' = \frac{a}{(1+x)^2}
\frac{a}{(1+x)^2}=-1
(1+x)^2=-a
x=\pm \sqrt{-a}-1
y = \frac{a(\pm \sqrt{-a}-1)}{\pm \sqrt{-a}}
Equation of tangent:
y = -(x-3)
x+y=3
(\pm \sqrt{-a}-1) + \frac{a(\pm \sqrt{-a}-1)}{\pm \sqrt{-a}} = 3
\pm a\mp \sqrt{-a} \pm a\sqrt{-a}-a = \pm 3\sqrt{-a}
Take +ve sign first
a - \sqrt{-a} + a\sqrt{-a}-a = 3\sqrt{-a}
a\sqrt{-a} = 4\sqrt{-a}
a = 4 Not possible
Take -ve sign
-a+\sqrt{-a}-a\sqrt{-a}-a = -3\sqrt{-a}
-a\sqrt{-a}-2a = -4\sqrt{-a}
a\sqrt{-a}+2a = 4\sqrt{-a}
Third and 4thoption are not clear. Solve this to get answer.

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