Question

Practice set 1( mathamatics)
Important for jee mains exam ​

Answer 1

Answer:

Option (a) is correct

Step-by-step explanation:

Given line ax+by+c is tangent to the curve xy=4

so we have find and equate the slope of the line and curve

Curve, xy = 4

• → x•dy/dx + y•(1) = 0

• → dy/dx = -y/x

• → dy/dx = -4/x²

slope or the line ax+by+ c = 0 is -a/b

since , the given line is tangent to the curve

• ∴ -4/x² = -a/b

• → a/b > 0

which is possible only when a>0 ,b >0 or a<0, b<0

Option (1) is marching with our posiblities so a> 0 and b>0 is answer

Answer 2

Answer:

xy = 4

on differentiating w.r.t x , we get

$→y + x \frac{dy}{dx} = 0$

$→ \frac{dy}{dx} = \frac{ - y}{x}$

$→ \frac{dy}{dx} = \frac{ - \frac{4}{x} }{x}$ (putting y = 4/x from eqn of curve)

$→slope \: = \frac{ - 4}{ {x}^{2} }$

and

ax+by+c = 0 is a tangent to the cure xy = 4. so, the slope will be -a/b.

$→ \frac{ - a}{b} = \frac{ - 4}{ {x}^{2} }$

$→ \frac{a}{b} = \frac{4}{ {x}^{2} }$

$→ \frac{a}{b} > 0$

which is possible only when a>0 and b>0 or a<0 and b<0.

Hence,(a) is correct.