Question

The equation of the normal drawn at the point \((\sqrt{2}+1, -1)\) to the ellipse \(x^2 + 2y^2 - 2x + 8y + 5 = 0\) is

• A. \(x+y=\sqrt{2} \)
• B. \(x-2y=3+\sqrt{2} \)
• C. \(\sqrt{2}x-y=3+\sqrt{2} \)
• D. \(2x+y=2\sqrt{2}+1 \)

Hint:

To find the equation of a normal to a curve \(F(x,y)=0\) at a point \((x_0, y_0)\): 1. Verify that the point lies on the curve. 2. Find the slope of the tangent \((m_T)\) by implicit differentiation \(\left(\frac{dy}{dx}\right)\) at \((x_0, y_0)\) or using partial derivatives \(\left(-\frac{\partial F / \partial x}{\partial F / \partial y}\right)\). 3. Calculate the slope of the normal \((m_N)\) using the perpendicularity condition: \(m_N = -\frac{1}{m_T}\) (if \(m_T \ne 0\)). If \(m_T = 0\), the tangent is horizontal and the normal is vertical (\(x = x_0\)). If \(m_T\) is undefined, the tangent is vertical and the normal is horizontal (\(y = y_0\)). 4. Use the point-slope form \(y - y_0 = m_N(x - x_0)\) to write the equation of the normal.

Answer 1

The Correct Option is C

Step 1: Verify if the given point lies on the ellipse.
The equation of the ellipse is \(x^2 + 2y^2 - 2x + 8y + 5 = 0\).
The given point is \(P(\sqrt{2}+1, -1)\).
Substitute \(x = \sqrt{2}+1\) and \(y = -1\) into the equation: \[ (\sqrt{2}+1)^2 + 2(-1)^2 - 2(\sqrt{2}+1) + 8(-1) + 5 \] \[ = (2 + 1 + 2\sqrt{2}) + 2(1) - (2\sqrt{2} + 2) - 8 + 5 \] \[ = 3 + 2\sqrt{2} + 2 - 2\sqrt{2} - 2 - 8 + 5 \] \[ = (3 + 2 - 2 - 8 + 5) + (2\sqrt{2} - 2\sqrt{2}) \] \[ = (10 - 10) + 0 = 0 \] Since the equation holds true, the point \(P(\sqrt{2}+1, -1)\) lies on the ellipse. 
Step 2: Find the slope of the tangent to the ellipse at the given point using implicit differentiation.
Differentiate the equation \(x^2 + 2y^2 - 2x + 8y + 5 = 0\) implicitly with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(2y^2) - \frac{d}{dx}(2x) + \frac{d}{dx}(8y) + \frac{d}{dx}(5) = 0 \] \[ 2x + 4y \frac{dy}{dx} - 2 + 8 \frac{dy}{dx} + 0 = 0 \] Group terms with \(\frac{dy}{dx}\): \[ (4y + 8) \frac{dy}{dx} = 2 - 2x \] \[ \frac{dy}{dx} = \frac{2 - 2x}{4y + 8} = \frac{2(1 - x)}{4(y + 2)} = \frac{1 - x}{2(y + 2)} \] This is the slope of the tangent, \(m_T\). Now, substitute the coordinates of the point \(P(\sqrt{2}+1, -1)\) into the expression for \(\frac{dy}{dx}\):
\[ m_T = \frac{1 - (\sqrt{2}+1)}{2(-1 + 2)} \] \[ m_T = \frac{1 - \sqrt{2} - 1}{2(1)} \] \[ m_T = \frac{-\sqrt{2}}{2} \] Step 3: Find the slope of the normal.
The normal to a curve at a point is perpendicular to the tangent at that point. If \(m_T\) is the slope of the tangent and \(m_N\) is the slope of the normal, then \(m_N \cdot m_T = -1\). \[ m_N = -\frac{1}{m_T} = -\frac{1}{(-\sqrt{2}/2)} = \frac{2}{\sqrt{2}} = \sqrt{2} \] Step 4: Find the equation of the normal.
The equation of a line passing through a point \((x_0, y_0)\) with slope \(m\) is given by \(y - y_0 = m(x - x_0)\). Using the point \(P(\sqrt{2}+1, -1)\) and slope \(m_N = \sqrt{2}\): \[ y - (-1) = \sqrt{2}(x - (\sqrt{2}+1)) \] \[ y + 1 = \sqrt{2}x - \sqrt{2}(\sqrt{2}) - \sqrt{2}(1) \] \[ y + 1 = \sqrt{2}x - 2 - \sqrt{2} \] Rearrange the equation to match the given options: \[ \sqrt{2}x - y = 1 + 2 + \sqrt{2} \] \[ \sqrt{2}x - y = 3 + \sqrt{2} \] The final answer is $\boxed{\sqrt{2}x-y=3+\sqrt{2}}$.