Question

If the normal drawn at a point \(P\) on the curve \[ 3y=6x-5x^3 \] passes through \((0,0)\), then the positive integral value of the abscissa of the point \(P\) is

• A. \(1\)
• B. \(\frac{2}{3}\)
• C. \(\frac{1}{3}\)
• D. \(-\frac{2}{3}\)

Hint:

For normal-related questions, first find the slope of tangent using differentiation, then use \[ m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}} \] and apply the condition that the normal passes through the given point.

Answer 1

The Correct Option is A

Step 1: Write the curve in simple form.
Given, \[ 3y=6x-5x^3 \] Dividing by \(3\), \[ y=2x-\frac{5}{3}x^3 \] Let the point \(P\) on the curve be \[ P(t,\ 2t-\frac{5}{3}t^3) \]

Step 2: Find the slope of tangent.
Differentiate \[ y=2x-\frac{5}{3}x^3 \] with respect to \(x\): \[ \frac{dy}{dx}=2-5x^2 \] At \(x=t\), slope of tangent is \[ m_t=2-5t^2 \] Therefore, slope of normal is \[ m_n=-\frac{1}{2-5t^2} \]

Step 3: Use the condition that normal passes through origin.
The normal passes through \[ (0,0) \] and \[ \left(t,\ 2t-\frac{5}{3}t^3\right) \] So, slope of the normal is also \[ m_n=\frac{2t-\frac{5}{3}t^3-0}{t-0} \] \[ m_n=2-\frac{5}{3}t^2 \] Thus, \[ 2-\frac{5}{3}t^2=-\frac{1}{2-5t^2} \]

Step 4: Solve for \(t\).
Cross multiplying, \[ \left(2-\frac{5}{3}t^2\right)(2-5t^2)=-1 \] Expanding, \[ 4-10t^2-\frac{10}{3}t^2+\frac{25}{3}t^4=-1 \] \[ 4-\frac{40}{3}t^2+\frac{25}{3}t^4=-1 \] Multiplying by \(3\), \[ 12-40t^2+25t^4=-3 \] \[ 25t^4-40t^2+15=0 \] Dividing by \(5\), \[ 5t^4-8t^2+3=0 \] \[ 5t^4-5t^2-3t^2+3=0 \] \[ 5t^2(t^2-1)-3(t^2-1)=0 \] \[ (t^2-1)(5t^2-3)=0 \] So, \[ t^2=1 \] or \[ t^2=\frac{3}{5} \] The positive integral value of \(t\) is \[ t=1 \]

Step 5: Final conclusion.
Therefore, the positive integral value of the abscissa of point \(P\) is \[ \boxed{1} \]