Question

Find the normals to the curve \(x y+2 x-y=0\) that are parallel to the line $2 x+y=0 .

Short answer

The 'x' coordinates at which the normals to the given curve are parallel to the specified line are the roots of the equation \(x^3 - x + 1 = 0\). The corresponding 'y' coordinates can be obtained by substituting these 'x' values into the equation \(y = 2 - 2x^2\).

Step-by-step solution

Find the derivative of the given curve

Differentiate the given curve equation with respect to \(x\). Applying the product rule on \(xy\) and differentiating \(2x\) and \(-y\) separately, we get:\(\frac{dy}{dx} = \frac{1-y-2}{x}\)

Find the slope of normals

The slope of the curve at a point is given by \(\frac{dy}{dx}\), so the slope of the normals is \(-\frac{1}{\frac{dy}{dx}}\). This follows because the usual rule applies that the product of the slopes of two perpendicular lines equals -1.

Equate the slope of normals to the slope of given line

The slope of the given line \(2x + y = 0\) is -2. Since the normals are parallel to this line, their slope also will be -2. So, \(-\frac{1}{\frac{dy}{dx}} = -2\). Substituting the value of \(\frac{dy}{dx}\) from Step 1, we obtain the equation \(2x^2 + y - 2 = 0\).

Incorporate this into the original equation of the curve

The existing equation of the curve is \(x y+2 x-y=0\). When we substitute \(y = 2 - 2x^2\) into this, we get, after simplifying, \(x^3 - x + 1 = 0\). The roots of this equation will give possible 'x' coordinates of the points where the normals exist.

Find the corresponding y-coordinates

When 'x' coordinates have been found, solve for 'y' by substituting the obtained 'x' values into the simplified equation from step 4 to find the corresponding y coordinates.