Question

The equation of the normal to the curve \(xy+7 = 0\) is \(Ax+By+C = 0\), then

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

Hint:

Find the slope of the tangent, then the slope of the normal, and compare it with the slope of the line Ax + By + C = 0.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The normal is perpendicular to the tangent. Its slope is the negative reciprocal of the tangent slope. A line \(Ax + By + C = 0\) has slope \(-A/B\).

Step 2: Tangent slope.
\(xy = -7\), so \(y = -\dfrac{7}{x}\) and \(\dfrac{dy}{dx} = \dfrac{7}{x^2}\). This is positive at every point of the curve (where \(x \neq 0\)).

Step 3: Normal slope.
\[ m_{normal} = -\frac{x^2}{7} < 0 \]

Step 4: Link to A and B.
\(-\dfrac{A}{B} < 0 \Rightarrow \dfrac{A}{B} > 0\), so A and B have the same sign. That means \(A > 0, B > 0\) or \(A < 0, B < 0\).

Step 5: Why the other options are wrong.
(B) and (C) say A and B have opposite signs, which would give a positive slope. (D) says \(C = 0\), which holds only if the normal passes through the origin, and that is not true for every normal.

Final Answer:
A and B have the same sign, option (A). \[ \boxed{A > 0,\ B > 0 \text{ or } A < 0,\ B < 0} \]