"Published in Newark, California, USA"
Find the equations of the tangent line and the normal line that passes through the point P(-3, -1) for the curve:
Solution:
The first thing that we have to do is to check if the given point is included in the curve. Consider the given equation of a curve
The given equation represents a conic section because both x and y are second degree but we cannot identify the exactly type of a conic section because of the xy term. To find out what type of a conic section is, we can use the discriminant formula as follows
where B is the coefficient of xy, A is the coefficient of x² and C is the coefficient of y². Substitute the values of A, B, C to the above equation, we have
Since the value of a discriminant is greater than zero, then the given conic section is hyperbola in which its transverse and conjugate axes are not parallel to x and y axes.
From the coordinates of the given point, substitute x = -3 and y = -1 at the given equation, we have
Since both sides of the equation are equal, then the given point is included in the curve.
Again, consider the given equation of a curve
Take the derivative on both sides of the equation with respect to x by implicit differentiation, we have
The slope of a curve is equal to the first derivative of the equation of a curve with respect to x. In this case, dy/dx is the slope of a curve.
To get the value of the slope of a curve at the given point, substitute x = 3 and y = 1 at the equation above, we have
The slope of a curve at the given point is equal to the slope of tangent line that passes thru also at the given point. Hence,
Therefore, the equation of a tangent line is
Normal line is also a straight line which is perpendicular to tangent line at the point of tangency. Hence,
Therefore, the equation of a normal line is
