Differential Calculus> Q. IF THE LINE JOINING THE POINTS (0,3) ...

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

Bharat Makkar , 11 Years ago

Grade Select Grade

1 Answers

Jitender Singh

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 4^thoption are not clear. Solve this to get answer.