Question

Find the equation of the tangent to the following parabola
3y^2+16x=0 ; two perpendicular tangents one of the two point of contact being (-3,4)

Answer 1

Hi!!

Answer 2

@SolomonZelman

Answer 3

Okay, I will reinterpret the question for you.

Answer 4

A horizontal parabola \(3y^2+16x=0\) will have the reflection of what is above the x-axis, below the x-axis. (Read this sentence carefully)

Thus, at x=-3 there are going to be two tangent lines.
And you have to choose the tangent line to the curve
at (-3,4) and (-3,-4), BUT you have to choose (-3,4).

Answer 5

So our function can be also written as:

\(3y^2+16x=0\)

\(3y^2=-16x\)

\(y^2=-\frac{16}{3}x\)

\(y=\color{red}{+} \sqrt{-\frac{16}{3}x}\)

Answer 6

I am choosing only the top half of the functio because the point that we are considering is on the top half of the function.

Answer 7

ok

Answer 8

So, \(f(x)= \sqrt{-\frac{16}{3}x}\)

find the equation of the tangent line, at the point (-3,4).

Answer 9

What would be the first thing to do here?

Answer 10

find the slope?

Answer 11

yes, and to do that you have to find the derivative.

Answer 12

What is the derivative of our function ?

Answer 13

y'=1/2*-16/3x^-1/2
y'=(-8/3)x^-1/2
y'=-8/3sq.rtx

Answer 14

correct yourself.

Answer 15

\[y'=\frac{ -8 }{ 3\sqrt{x} }\]

Answer 16

yes, but your quantity inside the square root is?

Answer 17

\(\color{#226633}{f(x)= \dfrac{-8}{3\sqrt{-\dfrac{16}{3}x}}}\)

Answer 18

Now, the slope at x=-3 is f\('\)(-3).
Can you find that ?

Answer 19

f(-3)=-2/3

Answer 20

yes, f\('\)(-3)=-2/3

Answer 21

https://www.desmos.com/calculator/9yd8w5xuyf