Question

The vertices of the ellipse given by
x2+4y2-2x=0 are

Answer 1

use the formula !

Answer 2

which formula

Answer 3

ok first of all you have to show the given equation into standar form !

Answer 4

like \[x^2/a^2+y^2/b^2 =1\]

Answer 5

or \[(x-h)^2/a^2 + (y-k)^2/b^2 =1 \]

Answer 6

(x-1) ^2/a^2+(2y-0)^2/b^2 = 1

Answer 7

can you show me how u got that equation !

Answer 8

x^2-2x+(2y) ^2=0
x^2-2x+1+(2y)^2=1

Answer 9

(x-1) ^ 2+(2y-0)^2=1

Answer 10

ok

Answer 11

then how u get a^2 and b^2

Answer 12

The base formula for ellipse is \[\frac{ (x-h) ^{2} }{ a^{2} } + \frac{ (y-k) ^{2} }{ b^{2} } = 1\]
Vertices of ellipse is f

Let see the given equation : \[x^{2}+4y^{2}-2x = 0\]
It is similar to : \[x^{2}-2x+4y^{2}=0\]
Let change it to: \[x^{2}-2x+1-1+4y^{2}=0\]

Change to right form: \[(x^{2}-2x+1) - 1 + 4y^{2}=0\]
It is similar to: \[(x-1)^{2} - 1 + 4y^{2}=0\]
then: \[(x-1)^{2} + 4y^{2}=1\]

Change it to be more closer to base ellipse formula:
\[\frac{ (x-1)^{2} }{ 1 } + \frac{ (y)^{2} }{ \frac{ 1 }{ 4 } } = 1\]

from the last form we got that the center (h,k) = (1,1)
the value of \[a^{2} = 1\], then a = \[\pm 1\]

The value of a also tells that the vertices are 1 units to either side of the center, at (h-a, k) and (h+a, k) which are (0,1) and (2,1)