Question

Consider the curves \( y = f(x) \) and \( x = g(y) \), and let \( P(x,y) \) be a common point of these curves. If at \( P \), on the curve \( y = f(x) \), \[ \frac{dy}{dx} = Q(x), \] and at the same point \( P \) on the curve \( x = g(y) \), \[ \frac{dx}{dy} = -Q(x), \] then:

• A. The two curves have a common tangent.
• B. The angle between two curves is \( 45^\circ \).
• C. The tangent drawn at \( P \) to one curve is normal to the other curve at \( P \).
• D. The two curves never intersect orthogonally.

Hint:

When checking orthogonality of two curves, compute the product of their slopes. If the product is \( -1 \), the curves are orthogonal.

Answer 1

The Correct Option is C

Step 1: Understanding the condition given 
For the curve \( y = f(x) \), the derivative at \( P(x, y) \) is: \[ \frac{dy}{dx} = Q(x). \] For the curve \( x = g(y) \), the derivative at \( P(x, y) \) is: \[ \frac{dx}{dy} = -Q(x). \] Using the chain rule: \[ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}. \] Substituting the given conditions: \[ \frac{dx}{dy} = \frac{1}{Q(x)}. \] 
Step 2: Check the product of the slopes 
If two curves are orthogonal, then the product of their slopes should be: \[ \frac{dy}{dx} \times \frac{dx}{dy} = -1. \] Substituting values: \[ Q(x) \times (-Q(x)) = - Q^2(x). \] \[ - Q^2(x) = -1. \] \[ Q^2(x) = 1. \] 

Step 3: Interpretation 
Since the product of slopes is \( -1 \), it means that the tangent at \( P \) to one curve is perpendicular to the tangent at \( P \) to the other curve. This implies that the tangent to one curve is normal to the other. 

Step 4: Conclusion 
Thus, the correct answer is: \[ \mathbf{\text{The tangent drawn at } P \text{ to one curve is normal to the other curve at } P.} \]