Question

Find the equations of the tangent and normal to each of the following conics, the lengths of the subtangent and subnormal, then trace the curve showing these lines.

y = x2 – 6x + 4 at (4, -4).

Answer 1

$$y=x^2-6x+4$$

$$y'=2x-6$$

Point $(4,-4)$

$$slope=m_1=y'(4)=2(4)-6=2$$

Tangent line is

$$y-(-4)=2(x-4)$$

The equation of the tangent line in slope-intercept form

$$y=2x-12$$

Find the equation of the normal at the point $(4, -4)$

$$slope=m_2=-\dfrac{1}{m_1}=-\dfrac{1}{2}$$

$$y-(-4)=-\dfrac{1}{2}(x-4)$$

The equation of the normal line in slope-intercept form

$$y=-\dfrac{1}{2}x-2$$

$TA$   is defined to be the subtangent at $P$. $AN$  is called the subnormal.

The lengths $PT$  and $PN$  are called the tangent and normal.

$$A(4, 0), T(6, 0), N(-4, 0)$$

$$PT=\sqrt{(6-4)^2+(0-(-4))^2}=2\sqrt{5}$$

$$PN=\sqrt{(-4-4)^2+(0-(-4))^2}=4\sqrt{5}$$

$$TA=2$$

$$AN=8$$