36x2 + 49y2 = 1,764 the foci are located at: a) (-√13, 0) and (√13,0) b) (0, -√13) and (0,√13) c) (-1, 0) and (1, 0)


4pounds of live bait and 1 pound of natural bait

i think it is 18

step-by-step explanation:

If y varies directly as x, and y is 180b when x is n when x is 5, what is the value of n? ​

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 \frac{x^2}{49}+\frac{y^2}{36}=1

This equation is of the form of ellipse whose standard equation can be written as \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

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


Substitute the values to find foci, we get

(0+\sqrt13,0) and (0-\sqrt13,0)

(\sqrt13,0) and (-\sqrt13,0)

Option (A) is correct.

Foci of given equation located at  (-√13, 0) and (√13,0)

Step-by-step explanation:

Consider the given equation 36x^2+49y^2=1764

We have to find the foci for the above equation,

First divide whole equation by 1764,

\Rightarrow \frac{36}{1764}x^2+\frac{49}{1764}y^2=\frac{1764}{1764}

\Rightarrow \frac{1}{49}x^2+\frac{1}{36}y^2=1

Since, both terms are positive so it is an equation of ellipse, a > b,

So it is a horizontal ellipse,

\frac{ (x-h)^2}{a^2}+ \frac{ (y-k)^2}{b^2}=1

With center at (h , k ) and foci ( h±c, k)  where, c^2=a^2-b^2

Given equation ,  \frac{1}{49}x^2+\frac{1}{36}y^2=1

 \frac{1}{7^2}x^2+\frac{1}{6^2}y^2=1 ,

Comparing a= 7 , b = 6 , h = k = 0

c^2=a^2-b^2=c^2=49-36=13 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.

Do you know the answer?

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