Applications of Derivatives
Normal to a Curve
Grade 12

Question:

<p>If the line <i>ax + by + c</i> = 0 is normal to the curve <i>xy</i> + 5 = 0, then <i>a</i> and <i>b</i> have</p>
<p>(a) same sign</p>
<p>(b) opposite sign</p>
<p>(c) cannot be discussed</p>
<p>(d) None of these</p>

Step-by-Step Solution

Key Concept: A line is normal to a curve if it is perpendicular to the tangent at the point of contact. For the curve xy + 5 = 0, we find the slope of the tangent using implicit differentiation, then use the condition that the normal's slope is the negative reciprocal of the tangent's slope to determine the relationship between coefficients a and b.
<p><strong>Step 1: Find the slope of the tangent to the curve xy + 5 = 0</strong></p><p>Differentiate implicitly with respect to x:</p><p>x(dy/dx) + y = 0</p><p>dy/dx = -y/x</p></p><p><strong>Step 2: Find the slope of the normal line</strong></p><p>The normal is perpendicular to the tangent, so:</p><p>Slope of normal = -1/(dy/dx) = -1/(-y/x) = x/y</p></p><p><strong>Step 3: Express the given line in slope form</strong></p><p>The line ax + by + c = 0 can be rewritten as:</p><p>by = -ax - c</p><p>y = -(a/b)x - (c/b)</p><p>Slope of the line = -a/b</p></p><p><strong>Step 4: Equate the slopes</strong></p><p>Since the line is normal to the curve:</p><p>-a/b = x/y</p><p>This must hold at the point of contact on the curve xy + 5 = 0.</p></p><p><strong>Step 5: Analyze the sign relationship</strong></p><p>From -a/b = x/y, we get: ay = -bx, or ay + bx = 0</p><p>Since the point (x, y) lies on xy + 5 = 0, we have xy = -5</p><p>This means x and y have opposite signs (one is positive, one is negative).</p><p>From ay + bx = 0 at the point of contact:</p><p>- If x > 0 and y < 0, then bx > 0 and ay < 0 (they must cancel), so b and a have opposite signs</p><p>- If x < 0 and y > 0, then bx < 0 and ay > 0 (they must cancel), so b and a have opposite signs</p></p><p>Therefore, a and b must have opposite signs.</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B

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