Let $L$ be a normal to the parabola $y^{2}=4x$. If $L$ passes through the point $(9, 6)$, then $L$ is given by

Let L be a normal to the parabola y²=4x. If L passes through the point (9, 6), then L is given by $\text{Given: } L \text{ is a normal to } y^2 = 4x$ $L \text{ passes through } (9,\, 6)$ $\text{To find: the equation of } L$ $\text{(from the four given options)}$ $1.\ \text{identify the parabola, plot the point}$ $2.\ \text{check the point lies on the parabola}$ $3.\ \text{tangent slope, then normal, then match}$ $y^2 = 4x \ \Rightarrow \ 4a = 4 \ \Rightarrow \ a = 1$ $\text{vertex } (0,0),\ \text{opens to the right}$ $(9,6):\quad y^2 = 36,\quad 4x = 36 \ \Rightarrow \ (9,6) \text{ lies on } y^2 = 4x$ $2y\,\dfrac{dy}{dx} = 4 \ \Rightarrow \ \dfrac{dy}{dx} = \dfrac{2}{y}$ $m_{\text{tangent}} = \dfrac{2}{6} = \dfrac{1}{3}$ $y - 6 = \dfrac{1}{3}(x - 9) \ \Rightarrow \ y = \dfrac{x}{3} + 3$ $m_{\text{normal}} = -\dfrac{1}{m_{\text{tangent}}} = -3$ $y - 6 = -3(x - 9)$ $y - 6 = -3x + 27 \ \Rightarrow \ y + 3x - 33 = 0$ $\text{check } (9,6):\ 6 + 3(9) - 33 = 0 \ \Rightarrow \ \text{verified}$ $\boxed{\text{Answer: (2)}\ \ y + 3x - 33 = 0}$