Question

At any point on the curve \[ by^2=(x+a)^3, \] if the length of the sub-tangent (ST) and the length of the sub-normal (SN) are such that \[ p(\text{SN})=q(\text{ST})^2, \] then \[ \frac{p}{q}= \]

• A. \(\dfrac{2b}{9}\)
• B. \(\dfrac{8b}{27}\)
• C. \(\dfrac{5b}{8}\)
• D. \(\dfrac{27}{8b}\)

Hint:

For tangent-normal problems, remember: \[ \text{Sub-tangent}=\frac{y}{dy/dx}, \qquad \text{Sub-normal}=y\frac{dy}{dx}. \] After finding \(dy/dx\), substitute directly into these formulas.

Answer 1

The Correct Option is B

Concept: For a curve \(y=f(x)\), \[ \text{Sub-tangent}=\frac{y}{\frac{dy}{dx}}, \] and \[ \text{Sub-normal}=y\frac{dy}{dx}. \] Use the equation of the curve to find \(\dfrac{dy}{dx}\), then evaluate ST and SN.

Step 1: Differentiate the given curve. Given, \[ by^2=(x+a)^3. \] Differentiating implicitly, \[ 2by\frac{dy}{dx} = 3(x+a)^2. \] Hence, \[ \frac{dy}{dx} = \frac{3(x+a)^2}{2by}. \] Using \[ by^2=(x+a)^3, \] \[ y^2=\frac{(x+a)^3}{b}. \] Therefore, \[ \frac{dy}{dx} = \frac{3(x+a)^2}{2(x+a)^{3/2}\sqrt b} = \frac{3}{2\sqrt b}\sqrt{x+a}. \]

Step 2: Find the sub-tangent. \[ \text{ST} = \frac{y}{\frac{dy}{dx}}. \] Using \[ y=\frac{(x+a)^{3/2}}{\sqrt b}, \] \[ \text{ST} = \frac{\frac{(x+a)^{3/2}}{\sqrt b}} {\frac{3}{2\sqrt b}\sqrt{x+a}}. \] \[ = \frac{2}{3}(x+a). \]

Step 3: Find the sub-normal. \[ \text{SN} = y\frac{dy}{dx}. \] \[ = \frac{(x+a)^{3/2}}{\sqrt b} \cdot \frac{3}{2\sqrt b}\sqrt{x+a}. \] \[ = \frac{3(x+a)^2}{2b}. \]

Step 4: Use the given relation. Given, \[ p(\text{SN}) = q(\text{ST})^2. \] Substituting ST and SN, \[ p\left(\frac{3(x+a)^2}{2b}\right) = q\left(\frac{2(x+a)}{3}\right)^2. \] \[ p\left(\frac{3(x+a)^2}{2b}\right) = q\left(\frac{4(x+a)^2}{9}\right). \] Cancelling \((x+a)^2\), \[ \frac{3p}{2b} = \frac{4q}{9}. \] \[ 27p=8bq. \] Hence, \[ \frac{p}{q} = \frac{8b}{27}. \] Therefore, \[ \boxed{\frac{p}{q}=\frac{8b}{27}} \] \[ \boxed{\text{Answer = (B)}} \]