Respuesta :
Answer:
Option (A) is correct.
Foci of given equation located at (-√13, 0) and (√13,0)
Step-by-step explanation:
Consider the given equation [tex]36x^2+49y^2=1764[/tex]
We have to find the foci for the above equation,
First divide whole equation by 1764,
[tex]\Rightarrow \frac{36}{1764}x^2+\frac{49}{1764}y^2=\frac{1764}{1764}[/tex]
[tex]\Rightarrow \frac{1}{49}x^2+\frac{1}{36}y^2=1[/tex]
Since, both terms are positive so it is an equation of ellipse, a > b,
So it is a horizontal ellipse,
[tex]\frac{ (x-h)^2}{a^2}+ \frac{ (y-k)^2}{b^2}=1[/tex]
With center at (h , k ) and foci ( h±c, k) where, [tex]c^2=a^2-b^2[/tex]
Given equation , [tex] \frac{1}{49}x^2+\frac{1}{36}y^2=1[/tex]
[tex] \frac{1}{7^2}x^2+\frac{1}{6^2}y^2=1[/tex] ,
Comparing a= 7 , b = 6 , h = k = 0
[tex]c^2=a^2-b^2=c^2=49-36=13[/tex] thus c = ±√13
Thus, foci is ( h±c, k) = (0 ±√13 ,0)
Thus, Foci of given equation located at (-√13, 0) and (√13,0)
Thus, option (A) is correct.
Answer:
The answer is A) (-√13, 0) and (√13,0)
Step-by-step explanation:
Given equation is 36x^2 + 49y^2 = 1,764
We have to find the foci
The above equation can be written as [tex]\frac{x^2}{49}+\frac{y^2}{36}=1[/tex]
This equation is of the form of ellipse whose standard equation can be written as [tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
where a represents the radius of the major axis and b represents the radius of the minor axis of the ellipse.
By comparing the above two equations, we get
a=7, b=6, h=0 and k=0
The foci of ellipse are (h+c,k) and (h-c,k)
c is the distance from the center to a focus which can be calculated as
[tex]\sqrt(a^2-b^2)=\sqrt(7^2-6^2)=\sqrt13[/tex]
Substitute the values to find foci, we get
[tex](0+\sqrt13,0)[/tex] and [tex](0-\sqrt13,0)[/tex]
⇒ [tex](\sqrt13,0)[/tex] and [tex](-\sqrt13,0)[/tex]
