Let $\ell$ be the tangent line to the ellipse $x^2 + 16y^2 = 4$ at $\left(1, \frac{\sqrt{3}}{4}\right)$. What is the equation of the line perpendicular to $\ell$ passing through $(2, 0)$?
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For any curve $f(x, y) = c$, the slope of the normal (which is perpendicular to the tangent) at $(x_1, y_1)$ is simply:
\[ m_{\text{normal}} = \frac{\partial f / \partial y}{\partial f / \partial x} \]
Evaluating at $\left(1, \frac{\sqrt{3}}{4}\right)$ gives $\frac{32y}{2x} = 16 \left(\frac{\sqrt{3}}{4}\right) = 4\sqrt{3}$ directly.
Step 1 : Understanding the Question:
The question asks us to find the equation of a line perpendicular to the tangent line of the given ellipse at a specific point, passing through $(2, 0)$. Step 2 : Key Formulas and Approach:
• We will find the slope of the tangent line ($m_{\text{tangent}}$) by differentiating the equation of the ellipse $x^2 + 16y^2 = 4$.
• The slope of the perpendicular line ($m_{\text{perp}}$) is given by the relation:
\[ m_{\text{perp}} = -\frac{1}{m_{\text{tangent}}} \]
• We will write the equation of the line using the point-slope form:
\[ y - y_1 = m_{\text{perp}}(x - x_1) \] Step 3 : Detailed Explanation:
Let us differentiate the equation of the ellipse $x^2 + 16y^2 = 4$ implicitly with respect to $x$:
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(16y^2) = \frac{d}{dx}(4) \]
\[ 2x + 32y \frac{dy}{dx} = 0 \]
\[ \frac{dy}{dx} = -\frac{2x}{32y} = -\frac{x}{16y} \]
Now, let us find the slope of the tangent line at the point $\left(1, \frac{\sqrt{3}}{4}\right)$:
\[ m_{\text{tangent}} = \left. \frac{dy}{dx} \right|_{\left(1, \frac{\sqrt{3}}{4}\right)} = -\frac{1}{16 \left(\frac{\sqrt{3}}{4}\right)} = -\frac{1}{4\sqrt{3}} \]
The slope of the line perpendicular to this tangent is:
\[ m_{\text{perp}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-\frac{1}{4\sqrt{3}}} = 4\sqrt{3} \]
Using the point-slope form for a line passing through $(2, 0)$ with slope $4\sqrt{3}$:
\[ y - 0 = 4\sqrt{3}(x - 2) \]
\[ y = 4\sqrt{3}(x - 2) \] Step 4 : Final Answer:
The equation of the perpendicular line is $y = 4\sqrt{3}(x - 2)$.
This corresponds to Option (A).
