Question

If the normal drawn at the point \((2, -1)\) to the ellipse \(x^2 + 4y^2 = 8\) meets the ellipse again at \((a, b)\), then \(17a\) is:

• A. 23
• B. 14
• C. 37
• D. 9

Hint:

- When dealing with tangents and normals to conic sections, use the properties of the derivatives to find the slope of the tangent and normal at any point. Then, apply the normal equation and solve for the intersections.

Answer 1

The Correct Option is B

Step 1: Find the slope of the tangent at the point \((2, -1)\). 
- The given ellipse equation is: \[ x^2 + 4y^2 = 8 \] - Differentiate both sides implicitly with respect to \(x\): \[ 2x + 8y \frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{x}{4y} \] - Substitute the coordinates \((2, -1)\) into the equation: \[ \frac{dy}{dx} = -\frac{2}{4(-1)} = \frac{1}{2} \] 

Step 2: Find the slope of the normal. 
- The slope of the normal is the negative reciprocal of the slope of the tangent: \[ \text{slope of normal} = -\frac{1}{\frac{1}{2}} = -2 \] 

Step 3: Find the equation of the normal. 
- The normal passes through the point \((2, -1)\) and has a slope of \(-2\). Using the point-slope form of the equation of a line: \[ y - (-1) = -2(x - 2) \] \[ y + 1 = -2(x - 2) \] \[ y + 1 = -2x + 4 \] \[ y = -2x + 3 \]

 Step 4: Substitute the equation of the normal into the ellipse equation. 
- The equation of the ellipse is: \[ x^2 + 4y^2 = 8 \] - Substitute \(y = -2x + 3\) into this equation: \[ x^2 + 4(-2x + 3)^2 = 8 \] - Expand \((-2x + 3)^2\): \[ x^2 + 4(4x^2 - 12x + 9) = 8 \] \[ x^2 + 16x^2 - 48x + 36 = 8 \] \[ 17x^2 - 48x + 36 = 8 \] \[ 17x^2 - 48x + 28 = 0 \] 

Step 5: Solve the quadratic equation. 
- Solve \(17x^2 - 48x + 28 = 0\) using the quadratic formula: \[ x = \frac{-(-48) \pm \sqrt{(-48)^2 - 4(17)(28)}}{2(17)} \] \[ x = \frac{48 \pm \sqrt{2304 - 1904}}{34} \] \[ x = \frac{48 \pm \sqrt{400}}{34} \] \[ x = \frac{48 \pm 20}{34} \] - This gives two solutions for \(x\): \[ x = \frac{48 + 20}{34} = \frac{68}{34} = 2 \quad \text{or} \quad x = \frac{48 - 20}{34} = \frac{28}{34} = \frac{14}{17} \]

 Step 6: Find the corresponding \(y\)-coordinates. 
- For \(x = 2\), \(y = -2(2) + 3 = -4 + 3 = -1\), which corresponds to the original point \((2, -1)\). - For \(x = \frac{14}{17}\), substitute into \(y = -2x + 3\): \[ y = -2\left(\frac{14}{17}\right) + 3 = -\frac{28}{17} + \frac{51}{17} = \frac{23}{17} \] 

Step 7: Find \(a\) and calculate \(17a\). 
- The new point of intersection is \(\left(\frac{14}{17}, \frac{23}{17}\right)\), so \(a = \frac{14}{17}\). - Calculate \(17a\): \[ 17a = 17 \times \frac{14}{17} = 14 \] Thus, \(17a = 14\).