Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

What is the center of the ellipse x2+36y2−72=0x^2 + 36y^2 - 72 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Find properties of ellipses from equations in general formright chevron icon

Full solution

Q. What is the center of the ellipse x2+36y2−72=0x^2 + 36y^2 - 72 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide by 7272: Divide the entire equation by 7272 to get the standard form of the ellipse equation.\newlinex272+36y272=7272\frac{x^2}{72} + \frac{36y^2}{72} = \frac{72}{72}\newlinex272+y22=1\frac{x^2}{72} + \frac{y^2}{2} = 1
  2. Identify center: Identify the center of the ellipse by comparing the equation to the standard form (x−h)2/a2+(y−k)2/b2=1(x-h)^2/a^2 + (y-k)^2/b^2 = 1. Here, hh and kk are both 00, so the center is at (0,0)(0, 0).

More problems from Find properties of ellipses from equations in general form

Question
Write the equation in standard form for the ellipse with center at the origin, vertex (−10,0)(-10,0), and co-vertex (0,3)(0,3).
Get tutor helpright-arrow

Posted 1 month ago

Question
Write the equation in standard form for the ellipse 7x2+y2=217x^2 + y^2 = 21.\newline______
Get tutor helpright-arrow

Posted 2 months ago

Question
What is the center of the ellipse x2+2y2−16=0x^2 + 2y^2 - 16 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(_____,_____)(\_\_\_\_\_,\_\_\_\_\_)
Get tutor helpright-arrow

Posted 1 month ago

Question
What is the length of the major axis of the ellipse x29+y22=1\frac{x^2}{9} + \frac{y^2}{2} = 1?\newlineWrite your answer in simplified, rationalized form.\newline_____\_\_\_\_\_
Get tutor helpright-arrow

Posted 1 month ago

Question
What is the center of the ellipse ((x−5)29)+((y−2)254)=1\left(\frac{(x - 5)^2}{9}\right) + \left(\frac{(y - 2)^2}{54}\right) = 1?\newlineWrite your answer in simplified, rationalized form.\newline(________ , _______)
Get tutor helpright-arrow

Posted 2 months ago

Question
The equation of an ellipse is given below.\newline(x+15)2676+(y−4)2100=1 \frac{(x+15)^{2}}{676}+\frac{(y-4)^{2}}{100}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (−15+24,4) (-15+\sqrt{24}, 4) and (−15−24,4) (-15-\sqrt{24}, 4) \newline(B) (−15,4+24) (-15,4+\sqrt{24}) and (−15,4−24) (-15,4-\sqrt{24}) \newline(C) (−39,4) (-39,4) and (9,4) (9,4) \newline(D) (−15,28) (-15,28) and (−15,−20) (-15,-20)
Get tutor helpright-arrow

Posted 3 months ago

Question
The equation of an ellipse is given below.\newline(x−5)23+(y−7)26=1 \frac{(x-5)^{2}}{3}+\frac{(y-7)^{2}}{6}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (5,7+3) (5,7+\sqrt{3}) and (5,7−3) (5,7-\sqrt{3}) \newline(B) (−5,−7+3) (-5,-7+\sqrt{3}) and (−5,−7−3) (-5,-7-\sqrt{3}) \newline(C) (−5+3,−7) (-5+\sqrt{3},-7) and (−5−3,−7) (-5-\sqrt{3},-7) \newline(D) (5+3,7) (5+\sqrt{3}, 7) and (5−3,7) (5-\sqrt{3}, 7)
Get tutor helpright-arrow

Posted 3 months ago

Question
The equation of an ellipse is given below.\newlinex246+(y+8)226=1 \frac{x^{2}}{46}+\frac{(y+8)^{2}}{26}=1 \newlineWhat are the foci of this ellipse?\newlineChoose 11 answer:\newline(A) (0+20,8) (0+\sqrt{20}, 8) and (0−20,8) (0-\sqrt{20}, 8) \newline(B) (0,8+20) (0,8+\sqrt{20}) and (0,8−20) (0,8-\sqrt{20}) \newline(C) (0,−8+20) (0,-8+\sqrt{20}) and (0,−8−20) (0,-8-\sqrt{20}) \newline(D) (0+20,−8) (0+\sqrt{20},-8) and (0−20,−8) (0-\sqrt{20},-8)
Get tutor helpright-arrow

Posted 3 months ago

Question
Write an equation for an ellipse centered at the origin, which has foci at (±12,0) ( \pm 12,0) and vertices at (±13,0) ( \pm 13,0) .
Get tutor helpright-arrow

Posted 3 months ago

Question
Write an equation for an ellipse centered at the origin, which has foci at (0,±6) (0, \pm 6) and vertices at (0,±37) (0, \pm \sqrt{37}) .
Get tutor helpright-arrow

Posted 3 months ago

View all (54)