Answer to Question #271848 in Analytic Geometry for Bebe

Question #271848


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).


1
Expert's answer
2021-11-30T13:07:14-0500
y=x26x+4y=x^2-6x+4

y=2x6y'=2x-6


Point (4,4)(4,-4)

slope=m1=y(4)=2(4)6=2slope=m_1=y'(4)=2(4)-6=2


Tangent line is


y(4)=2(x4)y-(-4)=2(x-4)

The equation of the tangent line in slope-intercept form


y=2x12y=2x-12



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


slope=m2=1m1=12slope=m_2=-\dfrac{1}{m_1}=-\dfrac{1}{2}

y(4)=12(x4)y-(-4)=-\dfrac{1}{2}(x-4)

The equation of the normal line in slope-intercept form


y=12x2y=-\dfrac{1}{2}x-2

TATA   is defined to be the subtangent at PP. ANAN  is called the subnormal.

The lengths PTPT  and PNPN  are called the tangent and normal.


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


PT=(64)2+(0(4))2=25PT=\sqrt{(6-4)^2+(0-(-4))^2}=2\sqrt{5}

PN=(44)2+(0(4))2=45PN=\sqrt{(-4-4)^2+(0-(-4))^2}=4\sqrt{5}

TA=2TA=2

AN=8AN=8




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