Question

Determine an equation of the tangent line and the normal line at the given point on the curve.

x²+x y-y²=19 ;(5,6)

Answer 1

The equations of the tangent line and normal line at the given point (5,6) on the curve x²+xy-y²=19 are y - 6 = -4/7(x - 5) and y - 6 = 7/4(x - 5), respectively.

Step-by-step explanation:

To determine an equation of the tangent line and the normal line at the given point on the curve x²+x y-y²=19, we first need to calculate the derivative to find the slope of the tangent line at the point (5,6). To find the slope of the tangent, we differentiate implicitly with respect to x:

2x + y + x(dy/dx) - 2y(dy/dx) = 0

Now, we solve for dy/dx, which gives us the slope of the tangent line:

m_tangent = dy/dx = (2x - y) / (x - 2y)

Plugging in the point (5,6), we find the slope of the tangent line to be:

m_tangent = (2*5 - 6) / (5 - 2*6) = 4 / (-7)

The equation of the tangent line is y - y1 = m(x - x1), so:

y - 6 = -4/7(x - 5)

Next, we find the slope of the normal line, which is the negative reciprocal of the tangent's slope:

m_normal = -1/m_tangent = 7/4

The equation of the normal line, therefore, is y - y1 = m_normal(x - x1), so:

y - 6 = 7/4(x - 5)

Thus, the equations of the tangent line and the normal line at the point (5,6) are:

y - 6 = -4/7(x - 5)

y - 6 = 7/4(x - 5), respectively.