Question

The equation of the normal to the curve \(3x^2 + y^2 = 8\), which is parallel to the line \(x + 3y = 10\) is?

Answer 1

Step 1: Find the Gradient of the Given Line
The given line is \(x + 3y = 10\). Rearranging to slope-intercept form \(y = mx + c\), we get: \[ y = -\frac{1}{3}x + \frac{10}{3} \] Therefore, the gradient of the line is \(m = -\frac{1}{3}\).

Step 2: Differentiate the Curve Implicitly
The given curve is \(3x^2 + y^2 = 8\). Differentiating implicitly with respect to \(x\), we get: \[ 6x + 2y \frac{dy}{dx} = 0 \] Solving for \(\frac{dy}{dx}\), we find: \[ \frac{dy}{dx} = -\frac{3x}{y} \] This is the gradient of the tangent to the curve.

Step 3: Gradient of the Normal
The normal to the curve is perpendicular to the tangent. Thus, the gradient of the normal is the negative reciprocal of the tangent's gradient: \[ \text{Gradient of normal} = \frac{y}{3x} \]

Step 4: Use the Parallel Condition
The normal is parallel to the given line, which has a gradient of \(-\frac{1}{3}\). Therefore, the gradient of the normal must also be \(\frac{1}{3}\), leading to: \[ \frac{y}{3x} = 3 \] This gives: \[ y = 9x \]

Step 5: Find the Point of Intersection
Substitute \(y = 9x\) into the equation of the curve: \[ 3x^2 + (9x)^2 = 8 \] Simplifying: \[ 3x^2 + 81x^2 = 8 \quad \Rightarrow \quad 84x^2 = 8 \quad \Rightarrow \quad x^2 = \frac{2}{21} \] So: \[ x = \pm \sqrt{\frac{2}{21}}, \quad y = \pm 9\sqrt{\frac{2}{21}} \]

Final Answer: The points of intersection are \((x, y) = \left(\pm \sqrt{\frac{2}{21}}, \pm 9\sqrt{\frac{2}{21}}\right)\).