First, find the slope function of the curve by differentiating
implicitly with respect to .
Now rearrange the equation to make the subject.
Next find the slope of the given line by rearranging to make
The slope of this line is . Hence, the slope of a
line perpendicular to this is .
Now set the slope function for the curve equal to this value and
rearrange to form an equation connecting and .
Therefore, the - and -coordinates of the points on the curve
where the tangents are perpendicular to
the given line satisfy both this equation and the equation of the curve. The two equations can be solved simultaneously.
First, rearrange the linear equation to give
Now substitute this expression for into the equation of the
curve and solve the resulting quadratic equation for .
Now substitute each of these -values back into the linear
equation to find the -coordinate.
Therefore, the coordinates of the points on the curve where the
tangent is perpendicular to the given line are and