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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-\ta= 4/2(3)1/2\ta= 8/(3)1/2\ta= -1 +-( plus – minus) (3)1/2\t-2(plus – minus) 2(3)1/2

Bharat Makkar , 10 Years ago
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anser 1 Answers
Jitender Singh

Last Activity: 10 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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