If the curves y^2 = 12x - 3 and y^2 = 12 - kx | vedclass

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(B) Given curves are $y^2 = 12x - 3$ and $y^2 = 12 - kx$. Let the point of intersection be $(x_1, y_1)$. For $y^2 = 12x - 3$, differentiating with res

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$6$

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(B) Given curves are $y^2 = 12x - 3$ and $y^2 = 12 - kx$. Let the point of intersection be $(x_1, y_1)$. For $y^2 = 12x - 3$, differentiating with respect to $x$, we get $2y \frac{dy}{dx} = 12 \implies \frac{dy}{dx} = \frac{6}{y}$. Let $m_1 = \frac{6}{y_1}$. For $y^2 = 12 - kx$, differentiating with respect to $x$, we get $2y \frac{dy}{dx} = -k \implies \frac{dy}{dx} = -\frac{k}{2y}$. Let $m_2 = -\frac{k}{2y_1}$. Since the curves cut orthogonally, $m_1 m_2 = -1 \implies (\frac{6}{y_1})(-\frac{k}{2y_1}) = -1 \implies \frac{3k}{y_1^2} = 1 \implies y_1^2 = 3k$. At intersection point $(x_1, y_1)$, $12x_1 - 3 = 12 - kx_1 \implies x_1(12 + k) = 15 \implies x_1 = \frac{15}{12+k}$. Also $y_1^2 = 12 - kx_1 = 3k \implies 12 - k(\frac{15}{12+k}) = 3k \implies 12(12+k) - 15k = 3k(12+k) \implies 144 + 12k - 15k = 36k + 3k^2 \implies 3k^2 + 39k - 144 = 0 \implies k^2 + 13k - 48 = 0 \implies (k+16)(k-3) = 0$. Since $k > 0$, $k = 3$. The curve is $y^2 = 12 - 3x$. At $x = 1$, $y^2 = 12 - 3(1) = 9 \implies y = 3$ (taking $b=3$). The slope at $(1, 3)$ is $m = -\frac{3}{2(3)} = -\frac{1}{2}$. The length of the sub-tangent is $|\frac{y}{dy/dx}| = |\frac{3}{-1/2}| = 6$.

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

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