Question

If the curves \(y^2=12x-3\) and \(y^2=12-kx\) cut each other orthogonally then the length of the sub tangent at (1,b) on the curve \(y^2=12-kx\) is

• A. 4
• B. 6
• C. 5
• D. 12

Hint:

Remember the formulas for lengths related to tangents and normals. At point \((x_1, y_1)\) with slope m: - Length of Tangent: \( |\frac{y_1}{m}\sqrt{1+m^2}| \) - Length of Normal: \( |y_1\sqrt{1+m^2}| \) - Length of Sub-tangent: \( |\frac{y_1}{m}| \) - Length of Sub-normal: \( |y_1 m| \)

Answer 1

The Correct Option is B

Step 1: Find the condition for orthogonal intersection.
Differentiate both curves with respect to x.
For \(y^2 = 12x-3\): \( 2y \frac{dy}{dx} = 12 \implies m_1 = \frac{dy}{dx} = \frac{6}{y} \).
For \(y^2 = 12-kx\): \( 2y \frac{dy}{dx} = -k \implies m_2 = \frac{dy}{dx} = \frac{-k}{2y} \).
For orthogonal intersection, the product of the slopes must be -1.
\( m_1 m_2 = (\frac{6}{y})(\frac{-k}{2y}) = \frac{-6k}{2y^2} = \frac{-3k}{y^2} = -1 \).
This gives the condition \( y^2 = 3k \).
Step 2: Find the intersection point and the value of k.
At the intersection point, the y-coordinates are the same, so \( 12x-3 = 12-kx \).
Also, \(y^2 = 3k\). Substitute this into the first curve's equation: \( 3k = 12x-3 \implies 12x = 3k+3 \implies 4x = k+1 \).
Now substitute \(y^2=3k\) into the second curve's equation: \( 3k = 12-kx \).
From \(4x=k+1\), we have \(x = (k+1)/4\). Substitute this into the equation above.
\( 3k = 12 - k\frac{k+1}{4} \). Multiply by 4: \( 12k = 48 - k(k+1) = 48 - k^2 - k \).
\( k^2 + 13k - 48 = 0 \). Factoring gives \( (k+16)(k-3) = 0 \). So \(k=3\) or \(k=-16\). The problem seems to imply a single intersection point, which would be \(k=3\). Let's assume \(k=3\).
Step 3: Find the length of the sub-tangent.
The length of the sub-tangent at a point \((x_1, y_1)\) is given by \( |\frac{y_1}{dy/dx}| \).
The point is (1,b) on the curve \(y^2 = 12-kx\). With \(k=3\), the curve is \(y^2=12-3x\).
At \(x=1\), \(b^2 = 12-3(1) = 9 \implies b = \pm 3\).
The slope at \((1,b)\) is \( m = \frac{-k}{2y} = \frac{-3}{2b} \).
Length of sub-tangent = \( |\frac{b}{-3/(2b)}| = |\frac{2b^2}{-3}| = \frac{2(9)}{3} = 6 \).