The values of parameter such that the line ( log 2(1+5 a-a2)) x-5 y-(a2-5)=0 is a normal to the curve x y=1, may lie in the interval

Question

The values of parameter such that the line ( log 2(1+5 a-a2)) x-5 y-(a2-5)=0 is a normal to the curve x y=1, may lie in the interval (A) (-∞, 0)  (B) (5, ∞) (C) (0,5) (D) (-∞, 0) ∪(5, ∞)

Tardigrade Admin · Accepted Answer

x y=1 ⇒ y=(1/x) ⇒ (d y/d x)=-(1/x2) Slope of normal =x2>0,(x ≠ 0) Slope of given line ( log 2(1+5 a-a2)/5)>0 1+5 a-a2 >1 a2-5 a< 0 a ∈(0,5) .

Q. The values of parameter such that the line $(lo g_{2} (1 + 5 a - a^{2})) x - 5 y - (a^{2} - 5) = 0$ is a normal to the curve $x y = 1$ , may lie in the interval

$(- \infty, 0)$ $

(5, \infty)$

$(0, 5)$

$(- \infty, 0) \cup (5, \infty)$

$x y = 1 \Rightarrow y = \frac{1}{x} \Rightarrow \frac{d y}{d x} = - \frac{1}{x ^{2}}$
Slope of normal $= x^{2} > 0, (x \neq = 0)$
Slope of given line $\frac{l o g _{2} ( 1 + 5 a - a ^{2} )}{5} > 0$
$1 + 5 a - a^{2} > 1$
$a^{2} - 5 a < 0$
$a \in (0, 5) .$

Q. The values of parameter such that the line (log2​(1+5a−a2))x−5y−(a2−5)=0 is a normal to the curve xy=1, may lie in the interval

(−∞,0) $

(5, \infty)$

(0,5)

(−∞,0)∪(5,∞)

xy=1⇒y=x1​⇒dxdy​=−x21​ Slope of normal =x2>0,(x=0) Slope of given line 5log2​(1+5a−a2)​>0 1+5a−a2>1 a2−5a<0 a∈(0,5).

Q. The values of parameter such that the line $(lo g_{2} (1 + 5 a - a^{2})) x - 5 y - (a^{2} - 5) = 0$ is a normal to the curve $x y = 1$ , may lie in the interval

$(- \infty, 0)$ $

$(0, 5)$

$(- \infty, 0) \cup (5, \infty)$

$x y = 1 \Rightarrow y = \frac{1}{x} \Rightarrow \frac{d y}{d x} = - \frac{1}{x ^{2}}$
Slope of normal $= x^{2} > 0, (x \neq = 0)$
Slope of given line $\frac{l o g _{2} ( 1 + 5 a - a ^{2} )}{5} > 0$
$1 + 5 a - a^{2} > 1$
$a^{2} - 5 a < 0$
$a \in (0, 5) .$