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Find if a point is inside or outside of an ellipse

Print if a point is inside or outside of an ellipse
Ellipse equation
x2 + y2 + x + y + = 0
( x )2  +  ( y )2  = 1
2 2
Test point
Test point: ( , )
Point status:
Notes and input limit
           

Solved example

How to find if a point is inside or outside of an ellipse

Print find if a point is in or out
ellipse form: (x+h)^2/a^2 +(y+k)^2/b^2 =1

Notice that the addition of the two fractions at the left side of the ellipse equation are equal to 1. now we have to check what happens if the addition of the two fractions is less then 1.

Instead of one we write the variable n

(x+h)^2/a^2 +(y+k)^2/b^2 =n

Define a value  s  so that the multiplication of s and n is equal to 1 so:   s ‧ n = 1

The value of s should be: s=1/n
Now divide the equation of the ellipse by n to get: (x-h)^2/〖na〗^2 +(y-k)^2/(nb^2 )=1

Define new a,2 and b,2 these values are the new major and minor axes values, we see that

a,2 = na2         and         b,2 = nb2

We can easily see that the new major and minor values depend on the value of n, so when n is bigger then 1 the value of a, and b, are bigger, we can summarize the relations as follows:

n > 1
Point is outside the ellipse
n < 1
Point is inside the ellipse
n = 1
point is on the ellipse

Example − Find the location of a point relative to an ellipse

Print find if a point is in or out
An ellipse is given by the equation: (x-5)^2/16+(y+3)^2/9=1

Find the location of points   P(1 , −2)   and   P(1 , −1)   relative to the ellipse.

Ellipse

According to the ellipse definition

d1 + d2 is a constant and equal to 2a

The values of d1 and d2 are:

d_1=√((p_x-f_x1 )^2+(p_y-f_y1 )^2 )

d_2=√((p_x-f_x2 )^2+(p_y-f_y2 )^2 )

Now we can find the values of d1 + d2 and compare it to the value of 2a.

d1 + d2 > 2a
Point is outside the ellipse
d1 + d2 < 2a
Point is inside the ellipse
d1 + d2 = 2a
point is on the ellipse
First find the focus length by the equation: f=√(a^2-b^2 )=√(25-9)=4

And the foci coordinates are f1 and f2 are:

(fx1 , fy1) = (h − f , k) = (5 − 4 , −3) = (1 , −3)
(fx2 , fy2) = (h + f , k) = (5 + 4 , −3) = (9 , −3)

For point p(1, −2) we have

d_1=√((1-1)^2+(-2+3)^2 )=1
d_2=√((1-9)^2+(-2+3)^2 )=8.06
2a = 2 ∙ 5 = 10

d1 + d2 = 1 + 8.06 = 9.06

we see that   d1 + d2 < 2a   so the point is inside the ellipse.

The second point (1 , −1) will be analysed in another way. substitute the value of the point into the equation of the ellipse to get:

(1-5)^2/25+(-1+3)^2/9=16/25+4/9=1.08

Because   1.08   is bigger than 1 the point is outside of the ellipse.

We can realize that this method of checking the point location is easier.