Answer:
[tex] \rm \displaystyle y _{ \rm tangent} = - \frac{8}{5} x - \frac{5}{2} a[/tex]
[tex] \rm \displaystyle y _{ \rm normal} = \frac{5}{8} x - \frac{765}{128} a[/tex]
Step-by-step explanation:
we are given a equation of parabola and we want to find the equation of tangent and normal lines of the Parabola
finding the tangent line
equation of a line given by:
[tex] \displaystyle y = mx + b[/tex]
where:
- m is the slope
- b is the y-intercept
to find m take derivative In both sides of the equation of parabola
[tex] \displaystyle \frac{d}{dx} {y}^{2} = \frac{d}{dx} 16ax [/tex]
[tex] \displaystyle 2y\frac{dy}{dx}= 16a[/tex]
divide both sides by 2y:
[tex] \displaystyle \frac{dy}{dx}= \frac{16a}{2y}[/tex]
substitute the given value of y:
[tex] \displaystyle \frac{dy}{dx}= \frac{16a}{2( - 5a)}[/tex]
simplify:
[tex] \displaystyle \frac{dy}{dx}= - \frac{8}{5}[/tex]
therefore
[tex] \displaystyle m_{ \rm tangent} = - \frac{8}{5}[/tex]
now we need to figure out the x coordinate to do so we can use the Parabola equation
[tex] \displaystyle ( - 5a {)}^{2} = 16ax [/tex]
simplify:
[tex] \displaystyle x = \frac{25}{16} a[/tex]
we'll use point-slope form of linear equation to get the equation and to get so substitute what we got
[tex] \rm \displaystyle y - ( - 5a)= - \frac{8}{5} (x - \frac{25}{16} a)[/tex]
simplify which yields:
[tex] \rm \displaystyle y = - \frac{8}{5} x - \frac{5}{2} a[/tex]
finding the equation of the normal line
normal line has negative reciprocal slope of tangent line therefore
[tex] \displaystyle m_{ \rm normal} = \frac{5}{8}[/tex]
once again we'll use point-slope form of linear equation to get the equation and to get so substitute what we got
[tex] \rm \displaystyle y - ( - 5a)= \frac{5}{8} (x - \frac{25}{16} a)[/tex]
simplify which yields:
[tex] \rm \displaystyle y = \frac{5}{8} x - \frac{765}{128} a[/tex]
and we're done!
( please note that "a" can't be specified and for any value of "a" the equations fulfill the conditions)
