Question

If the line \(ax+by+5 = 0\) is a normal to the curve \(xy = 1\) then .....

• A. \(a > 0,b < 0\)
• B. \(a < 0,b < 0\)
• C. \(a > 0,b = 0\)
• D. \(a > 0,b > 0\)

Hint:

Slope of the normal to xy = 1 is x squared, which is always positive.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For the curve \(xy = 1\), \(y = \frac1x\), so \(\frac{dy}{dx} = -\frac{1}{x^2}\). The slope of the tangent is always negative.

Step 2: Slope of the normal:
The normal slope is the negative reciprocal: \(m = x^2 > 0\). So any normal has a positive slope (and cannot be vertical or horizontal).

Step 3: Compare with the line:
The line \(ax + by + 5 = 0\) has slope \(-\frac{a}{b}\), which must be positive. This needs \(a\) and \(b\) to have opposite signs, with \(b \ne 0\).

Step 4: Check the options:
(A) \(a > 0, b < 0\): opposite signs, so the slope is positive. This is possible.
(B) and (D) have same signs, giving a negative slope, which is impossible.
(C) \(b = 0\) gives a vertical line, which cannot be a normal here.

Final Answer:
Option (A) is correct. \[ \boxed{a>0,\ b<0} \]