Problem 68

Question

If the line $a x+b y+c=0$ is a normal to the hyperbola $x y=1$, then (A) $a>0, b<0$ (B) $a>0, b>0$ (C) $a<0, b<0$ (D) $a<0, b>0$

Step-by-Step Solution

Verified

Answer

(D) $a<0, b>0$.

1Step 1: Understand the Normal Line Condition

When a line $ ax + by + c = 0 $ is normal to a curve at any point, the slope of the line is the negative reciprocal of the derivative of the curve at that point.

2Step 2: Differentiate the Hyperbola Equation

The equation of the hyperbola given is $ xy = 1 $. To find the derivative $ \frac{dy}{dx} $, differentiate both sides with respect to $ x $. This gives $ y + x\frac{dy}{dx} = 0 $. Solving for $ \frac{dy}{dx} $ gives $ \frac{dy}{dx} = -\frac{y}{x} $.

3Step 3: Calculate the Slope of the Normal

If the line is normal to the hyperbola, its slope $ -\frac{a}{b} $ should be the negative reciprocal of the derivative $ -\frac{y}{x} $. This means $ -\frac{a}{b} = \frac{x}{y} $.

4Step 4: Analyze Slope Signs

For the hyperbola $ xy = 1 $, the product of the coordinates $ (x, y) $ is always positive. Therefore, if $ \frac{x}{y} $ is positive, then both $ x $ and $ y $ must have the same sign (both positive or both negative).

5Step 5: Derive Sign Conditions for $a$ and $b$

Since $ \frac{x}{y} $ corresponds to $ -\frac{a}{b} $, $ a $ and $ b $ must have opposite signs to maintain the negative slope condition. Thus, if $ \frac{x}{y} $ is positive, $ -\frac{a}{b} $ is positive, requiring $ a < 0 $ and $ b > 0 $.

6Step 6: Conclusion

Combining all the analyzed conditions, the correct assertion is that $ a < 0 $ and $ b > 0 $. Hence, the answer is option (D).

Key Concepts

Normal LineDifferentiationSlopeDerivative of Hyperbola

Normal Line

A normal line to a curve is a line that intersects the curve at a right-angle. It is effectively perpendicular to the tangent of the curve at that point of intersection.
This means if you know the slope of the tangent line, the normal line will have a slope that is the negative reciprocal of that of the tangent.
For example, if the slope of the tangent is $ m $, then the normal line would have a slope of $ -\frac{1}{m} $.

Understanding normal lines is essential in geometry as they help find the steepest point of descent from a curve.
This knowledge is useful in optimization problems or finding particular points related to curves.

Knowing about normal lines allows us to determine specific relationships and intersections between lines and curves through their slopes.

Differentiation

Differentiation is a mathematical process used to find the rate at which a function is changing at any given point.
It is the foundation for calculus and helps us understand the behavior of curves by providing the slope of the tangent line at any point along the curve.
When differentiating an equation like a hyperbola, we apply rules to systematically find the derivative.
For the hyperbola $ xy = 1 $, we find the derivative $ \frac{dy}{dx} $ by differentiating both sides with respect to $ x $.

This involves applying the product rule, resulting in $ y + x\frac{dy}{dx} = 0 $.
Solving gives $ \frac{dy}{dx} = -\frac{y}{x} $, which is crucial for understanding the curve's behavior.

Differentiation provides us with a powerful tool to analyze and predict the changes in variables as they relate to one another.

Slope

The slope is a measure of steepness or the degree of inclination of a line.
In the context of curves, the slope at a given point can be represented by the derivative of the function at that point.
For a line, the slope is often represented as $ \frac{rise}{run} $, which translates to $ \frac{change\ in\ y}{change\ in\ x} $.
With our hyperbola example, the slope of the normal line is derived from the condition that $ -\frac{a}{b} = \frac{x}{y} $.

This equation ensures that the line has the proper orientation relative to the curve, maintaining perpendicularity with the tangent.
This concept is crucial for determining angles of intersection and inclination between lines and curves.

Understanding slopes enriches our ability to assess the layout and orientation of curves and their respective lines.

Derivative of Hyperbola

The derivative of a hyperbola helps us understand how the curve changes at any point along it.
For the hyperbola given by $ xy = 1 $, finding the derivative involves using implicit differentiation.
By differentiating with respect to $ x $, we find $ y + x\frac{dy}{dx} = 0 $ and solve for $ \frac{dy}{dx} $.
The result, $ \frac{dy}{dx} = -\frac{y}{x} $, acts as a key in finding the rate of change for the curve.

This derivative describes the slope of the tangent line to any point on the hyperbola.
It allows us to solve for conditions where lines such as tangents and normals intersect the hyperbola.

Derivatives are powerful in giving insight into the dynamic properties of curves, determining concavity, and predicting future projection points.

Problem 67

Problem 69

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