Need help with tangent/normal derivative problem

[gloomy fog]

I need a walkthrough this problem

[gloomy fog]

Need help with tangent/normal derivative problem

[opaque zinc]

You know the first step when considering tangents/normal lines?

[gloomy fog]

I can plot 2 points on the curve, (x1,y1) and (x2, y2), then plug into the curve and get 2 equations and then find the remaining 2 equations using point-slope form and slope using derivative of the curve, but this approach is messy af so looking for a alternative

[opaque zinc]

I think that you can also start by differentiating both sides of the equation wrt x

[severe patrol]

is the answer tanh (1)

[gloomy fog]

the answer is +- sqrt 2

[severe patrol]

well

you see i saw the graph

and i dont see any where SL is a both a tangent and a normal

see i can show u how to solve it

u differentiate both sides

you get

27 * 2 y y' = 4*3 x^2

now y ' = 2x^2/9y

this y' is the slope of the tangent at any point (x,y)

say there exists (h,k) whihc satisfies ur condition

then m = 2h^2/9k

and we know that the slope of normal is (-1/slope of tangent)

whihc gives

slope of normal (say m') = - 9k/2h^2

now m m' = -1

and we also know that teh slope of normal must be equal to the slope of tangent

as its in the quesiton

"the slope of sl whihc is both a tangetn and normal"

which leaves us with

2h^2 /9k = -9k/2h^2

(4/81) h^4/k^2 = -1

whihc in its self is a false state ment

but we continue

we know that 4h^3= 27 k^2

whihc gives us k^2 = 4/27 h^3

[severe patrol]

when we substi k^2 in ts

we end up with

h = -3

now as h = -3 is not possible and does not lie on the curve

hecne no real slope exists

but lets continue anyways

now h = -3

[severe patrol]

which means k = 2 i (i = iota sqrt(-1))

now the m = 2h^2/9k

when we substi h,k

we end up with( -i ) as the slope

[severe patrol]

Mr

aint no fucking way i solved it to the air

omfg

[gloomy fog]

i did the same

what if we take parametric form of x and y?

[severe patrol]

idk try

but if it had to work

it would give soem real solutions

ohh

wiat wait

but

[gloomy fog]

x = 3t^2, and y = 2t^3, and plugging it into the derivative gives the slope as t

what to do next?

[severe patrol]

look at the them cutely

its not gon work

[opaque zinc]

that'll do it

so what you'll get is
$\frac{\text{d}y}{\text{d}x}=\frac{12(3t^2)^2)}{54(2t^3)}= \frac{12 \cdot 9t^4}{54 \cdot 2t^3} = t$

[eternal thistle]

Ties

[opaque zinc]

So an equation of the tangent at a point called P (P is the tangent point, so $P(3t^2, 2t^3))$
$$y-2t_1^3 = t_1(x-3t_1^2)$$
$$\Rightarrow t_1x-y = t_1^3$$

Equation of the normal at a point called Z will be
$$y-2t_2^3 = -\frac{1}{t_2} (x-3t_2^2)$$
$$\Rightarrow x+t_2 y = 2t_2^4 + 3t_2^2$$

[eternal thistle]

Ties

Ties

[opaque zinc]

We know that the tangent point and the normal point should be equal so,
$t_1= \frac{-1}{t_2} = \frac{t_1^3}{2t_2^4+3t_2^2}$

[eternal thistle]

Ties

[opaque zinc]

It follows $t_2 = -\frac{1}{t_1}$

[eternal thistle]

Ties