Let p be a positive constant and consider the parabola x^2=4py with vertex at the origin and focus at the point (0, p). (a) Show that the tangent line at (x_0, y_0) has y-intercept y_0.

tamkieuqf

tamkieuqf

Open question

2022-08-19

Let p be a positive constant and consider the parabola x 2 = 4 p y with vertex at the origin and focus at the point ( 0, p). (a) Show that the tangent line at ( x 0 , y 0 ) has y-intercept y 0 .
What I try:
1. the slope formula (i) 𝑦 y βˆ’ y 0 = m ( x βˆ’ x 0 )
2. m = d y d x x 2 4 p = x 2 p
Evaluated m when x = x 0 m = x 0 2 p
Replaced m in (i) y = x 0 2 p ( x βˆ’ x 0 ) + y 0
In order to get the y-intercept, I evaluated x = 0
y = x 0 2 2 p + y 0
What is missing?

Answer & Explanation

Irene Simon

Irene Simon

Beginner2022-08-20Added 16 answers

When x = 0, you'll get βˆ’ x 0 2 2 p + y 0 . Since ( x 0 , y 0 ) is a point on x Β² = 4 p y we have x 0 2 = 4 p y 0 .
The y-intercept is then
βˆ’ x 0 2 2 p + y 0 = βˆ’ 2 y 0 + y 0 = βˆ’ y 0
as desired.
trokusr

trokusr

Beginner2022-08-21Added 2 answers

Since the tangent line is
y = x 0 2 p ( x βˆ’ x 0 ) + y 0 ,
if you put x = 0, then you get
y = βˆ’ x 0 2 2 p + y 0 = βˆ’ x 0 2 2 p + x 0 2 4 p = βˆ’ x 0 2 4 p = βˆ’ y 0 .

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