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: Use implicit differentiation.
$xy + 7 = 0 \Rightarrow y + x\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{y}{x}$.

Step 2: Use the curve.
Because $xy = -7 < 0$, $x$ and $y$ have opposite signs, so $-y/x > 0$: the tangent slope is positive.

Step 3: Normal.
Normal slope is $x/y$, which is negative. For $Ax + By + C = 0$ the slope is $-A/B < 0$, so $A/B > 0$.

Final Answer:
Option (A). \[ \boxed{A > 0,\ B > 0 \text{ or } A < 0,\ B < 0} \]