Question

If \[ y=\frac{ax+b}{cx+d} \] and \[ \frac{dx}{dy}=\frac{ad-bc}{Py^2+Qy+R} \] then \(P+Q+R=\)

• A. \((a+c)^2\)
• B. \((a-c)^2\)
• C. \(\dfrac{ad-bc}{a^2+c^2-2ac}\)
• D. \(\dfrac{1}{(a-c)^2}\)

Hint:

For fractional linear functions, first express \(x\) in terms of \(y\), then differentiate carefully using the quotient rule.

Answer 1

The Correct Option is B

Step 1: Differentiate y directly using the quotient rule.
Given \(y=\frac{ax+b}{cx+d}\), the quotient rule gives
\[ \frac{dy}{dx} = \frac{a(cx+d) - c(ax+b)}{(cx+d)^2} = \frac{ad-bc}{(cx+d)^2} \]

Step 2: Write cx + d in terms of y.
From \(a - cy = a - c\cdot\frac{ax+b}{cx+d} = \frac{a(cx+d)-c(ax+b)}{cx+d} = \frac{ad-bc}{cx+d}\), we get
\[ cx+d = \frac{ad-bc}{a-cy} \]

Step 3: Substitute back and invert.
So \((cx+d)^2 = \frac{(ad-bc)^2}{(a-cy)^2}\), and taking the reciprocal of Step 1's result,
\[ \frac{dx}{dy} = \frac{(cx+d)^2}{ad-bc} = \frac{(ad-bc)^2/(a-cy)^2}{ad-bc} = \frac{ad-bc}{(a-cy)^2} \]

Step 4: Compare and expand.
Matching with \(\frac{dx}{dy}=\frac{ad-bc}{Py^2+Qy+R}\), we get \(Py^2+Qy+R=(a-cy)^2=c^2y^2-2acy+a^2\), so \(P=c^2, Q=-2ac, R=a^2\).

Step 5: Final conclusion.
\[ P+Q+R = a^2-2ac+c^2 = \boxed{(a-c)^2} \]