Question

Statement-I: The equation of the tangent to the curve \[ y=3x^2-5 \] drawn through the point \((1,2)\) is \[ y=6x-4. \] Statement-II: If \(L,M,N\) are respectively the lengths of tangent, normal and subnormal drawn to a curve at a point \((a,b)\), then \[ \frac{(L)(N)}{M}=b^2. \] Choose the correct option.

• A. Both statements I and II are correct
• B. Statement I is correct but statement II is not correct
• C. Statement I is not correct but statement II is correct
• D. Both statements I and II are not correct

Hint:

For statement-based questions, verify each statement independently before choosing the final option.

Answer 1

The Correct Option is A

Concept: To verify a tangent equation:

• Check whether the point lies on the curve.
• Compute the derivative.
• Verify the tangent line equation.

For tangent, normal and subnormal: \[ L=b\sqrt{1+m^2}, \] \[ M=b\sqrt{1+\frac1{m^2}}, \] \[ N=\frac{b}{m}, \] where \(m=\frac{dy}{dx}\).

Step 1: Verify Statement-I.
Given curve \[ y=3x^2-5. \] At \(x=1\), \[ y=3(1)^2-5=-2. \] The tangent at \(x=1\) passes through \[ (1,-2). \] Derivative: \[ \frac{dy}{dx}=6x. \] At \(x=1\), \[ m=6. \] Equation of tangent: \[ y+2=6(x-1). \] \[ y=6x-8. \] Hence the given equation \[ y=6x-4 \] is not the tangent. Therefore Statement-I is false.

Step 2: Verify Statement-II.
Using standard formulas, \[ L=b\sqrt{1+m^2}, \] \[ N=\frac{b}{m}, \] \[ M=\frac{b\sqrt{1+m^2}}{m}. \] Therefore \[ \frac{LN}{M} = \frac{ b\sqrt{1+m^2}\cdot\frac{b}{m} }{ \frac{b\sqrt{1+m^2}}{m} }. \] Cancelling common factors, \[ \frac{LN}{M}=b. \] Not \(b^2\). Hence Statement-II is also false. Therefore, \[ \boxed{\text{Both statements I and II are not correct}}. \]