How do you solve #3y^2-x^2=25#?

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Answer 1 · 2016-10-04T01:50:13

Vertex #(0,5) ; (0, -5)#

Covertex #(2.87, 0); (-2.87, 0)#

Foci #(0, 4.08); (0, -4.08)#

Given -

#3x^2+y^2=25#

#x^2/(25/3)+y^2/25=25/25#

#x^2/(25/3)+y^2/25=1#

Equation of the ellipse is -

#x^2/a+y^2/b=1#

Where

#a->#Major axis
#b-> #Minor axis

#a>b#
Then Major axis is along the X-axis

We have to take the bigger value as #a#
In our case, if we take 25 as #a#, Then major axis is Y-axis.

Then -

#a^2=25#
#b^2=25/3#

#a=sqrt25=5#
#b=sqrt(25/3)=2.87#

#c^2=a^2-b^2#
#c^2=25-(25/3)=(75-25)/3=50/3#

#c=sqrt(50/3)=4.08#

Vertex #(0,5) ; (0, -5)#

Covertex #(2.87, 0); (-2.87, 0)#

Foci #(0, 4.08); (0, -4.08)#

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