Question

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

Answer 1

`\rm \displaystyle y _( \rm tangent) = - (8)/(5) x - (5)/(2) a`

`\rm \displaystyle y _( \rm normal) = (5)/(8) x - (765)/(128) a`

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 (d)/(dx) {y}^(2) = (d)/(dx) 16ax`

`\displaystyle 2y(dy)/(dx)= 16a`

divide both sides by 2y:

`\displaystyle (dy)/(dx)= (16a)/(2y)`

substitute the given value of y:

`\displaystyle (dy)/(dx)= (16a)/(2( - 5a))`

simplify:

`\displaystyle (dy)/(dx)= - (8)/(5)`

therefore

`\displaystyle m_( \rm tangent) = - (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 = (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)= - (8)/(5) (x - (25)/(16) a)`

simplify which yields:

`\rm \displaystyle y = - (8)/(5) x - (5)/(2) a`

finding the equation of the normal line

normal line has negative reciprocal slope of tangent line therefore

`\displaystyle m_( \rm normal) = (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)= (5)/(8) (x - (25)/(16) a)`

simplify which yields:

`\rm \displaystyle y = (5)/(8) x - (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

In attachment

Explanation:

For answer refer to attachment .