Question

`find the equatifindon of tangent and normal of parabola y²=16ax at point whose coordinate is -5a. ​

Answer 1

Answer:

$\rm \displaystyle y _{ \rm tangent} = - \frac{8}{5} x - \frac{5}{2} a$

$\rm \displaystyle y _{ \rm normal} = \frac{5}{8} x - \frac{765}{128} a$

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:

$\displaystyle y = mx + b$

where:

• m is the slope
• b is the y-intercept

to find m take derivative In both sides of the equation of parabola

$\displaystyle \frac{d}{dx} {y}^{2} = \frac{d}{dx} 16ax$

$\displaystyle 2y\frac{dy}{dx}= 16a$

divide both sides by 2y:

$\displaystyle \frac{dy}{dx}= \frac{16a}{2y}$

substitute the given value of y:

$\displaystyle \frac{dy}{dx}= \frac{16a}{2( - 5a)}$

simplify:

$\displaystyle \frac{dy}{dx}= - \frac{8}{5}$

therefore

$\displaystyle m_{ \rm tangent} = - \frac{8}{5}$

now we need to figure out the x coordinate to do so we can use the Parabola equation

$\displaystyle ( - 5a {)}^{2} = 16ax$

simplify:

$\displaystyle x = \frac{25}{16} a$

we'll use point-slope form of linear equation to get the equation and to get so substitute what we got

$\rm \displaystyle y - ( - 5a)= - \frac{8}{5} (x - \frac{25}{16} a)$

simplify which yields:

$\rm \displaystyle y = - \frac{8}{5} x - \frac{5}{2} a$

finding the equation of the normal line

normal line has negative reciprocal slope of tangent line therefore

$\displaystyle m_{ \rm normal} = \frac{5}{8}$

once again we'll use point-slope form of linear equation to get the equation and to get so substitute what we got

$\rm \displaystyle y - ( - 5a)= \frac{5}{8} (x - \frac{25}{16} a)$

simplify which yields:

$\rm \displaystyle y = \frac{5}{8} x - \frac{765}{128} a$

and we're done!

( please note that "a" can't be specified and for any value of "a" the equations fulfill the conditions)

Answer 2

Answer:

In attachment

Step-by-step explanation:

For answer refer to attachment .