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

The equation of the normal to the curve \(3x^2 + y^2 = 8\), which is parallel to the line \(x + 3y = 10\), can be found by using the fact that the normal to a curve is perpendicular to the tangent at the point of intersection.

First, find the gradient of the given line by rearranging it in the slope-intercept form: \(y = (\frac{-1}{3})x + \frac{10}{3}\).

The gradient is \(\frac{-1}{3}\). Next, differentiate the given curve implicitly to find the gradient of the tangent at any point on the curve: \(\frac{dy}{dx} = -\frac{3x}{y}\).

For the normal to be parallel to the given line, the gradient of the tangent and the gradient of the normal must be negative reciprocals of each other.

Therefore, the gradient of the normal is \(\frac{3}{y}\). At the point of intersection of the curve and the line, the normal and tangent coincide.

Hence, substitute the coordinates of the point of intersection into the equation of the curve: \(3x^2 + y^2 = 8\).

Solving the simultaneous equations, \(x = 2\) and \(y = 2\).

Finally, using the point-slope form, the equation of the normal is \(y - 2 = (\frac{3}{2})(x - 2)\).

In brief, the equation of the normal to the curve \(3x^2 + y^2 = 8\), which is parallel to the line \(x + 3y = 10\), is \(y - 2 = (\frac{3}{2})(x - 2)\).