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# write an equation of an ellipse where

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write an equation of an ellipse where

the foci are at (1,-1) and (1,5) and the ellipse passes through (4,2) ???

Jan 21, 2018

#1
+18334
+1

It will have a vertical axis and the equation looks something like:

(x-1)^2/3^2 + (y-2)^2/4^2= 1      I'll have to do a little more review and get back to you....

The CENTER of the ellipse is directly between the foci

so center =   1,2    (h,k)

Equation for an ellipse looks like:

(x-h)^2 / b^2   +  (y-k)^2 / a^2 = 1    where h.k is the center (much like the equation for a circle)

so we have:

(x-1)^2 /a^2   + (y-2)^2 /b^2 = 1   we can find b^2 by substituting the point given (4,2)

results in b^2 = 9    or b =3

(x-1)^2/3^2 + (y-2)^2/a^2  = 1      Now we need to find a^2

The three letters are related by the equation b2 = a2 – c2 or, alternatively (depending on your book or instructor), by the equation b2 + c2 = a2.

(Proving the relationship requires pages and pages of algebraic computations, so just trust me that the equation is true. It can also be shown — painfully — that b is also the length of the semi-minor axis, so the distance across the ellipse in the "shorter" direction is 2b. Yes, the Pythagorean Theorem is involved in proving this stuff. Yes, these are the same letters used in the Pythagorean Theorem. No, this is not the same as the Pythagorean Theorem. Yes, this is very confusing. Accept it, make sure to memorize the relationship before the next test, and move on.)

OK   then      b^2 +c^2 = a^2       we know b = 3    c=3 (didstance from center to one of the foci) so we have:

3^2 +3^2 = a^2       thus a^2 = 18    and  FINALLY, we have the quation of the ellipse in question:

(x-1)^2 / 3^2   +  (y-2)^2 / 18  =1

(there PROBABY is an easier way !!!!   This was my thought process as I was reading through Purplemath learning about ellipses....you might want to read through it too ! )

Here is a desmos graph:

Jan 21, 2018
edited by ElectricPavlov  Jan 21, 2018
edited by ElectricPavlov  Jan 21, 2018
edited by ElectricPavlov  Jan 21, 2018

#1
+18334
+1

It will have a vertical axis and the equation looks something like:

(x-1)^2/3^2 + (y-2)^2/4^2= 1      I'll have to do a little more review and get back to you....

The CENTER of the ellipse is directly between the foci

so center =   1,2    (h,k)

Equation for an ellipse looks like:

(x-h)^2 / b^2   +  (y-k)^2 / a^2 = 1    where h.k is the center (much like the equation for a circle)

so we have:

(x-1)^2 /a^2   + (y-2)^2 /b^2 = 1   we can find b^2 by substituting the point given (4,2)

results in b^2 = 9    or b =3

(x-1)^2/3^2 + (y-2)^2/a^2  = 1      Now we need to find a^2

The three letters are related by the equation b2 = a2 – c2 or, alternatively (depending on your book or instructor), by the equation b2 + c2 = a2.

(Proving the relationship requires pages and pages of algebraic computations, so just trust me that the equation is true. It can also be shown — painfully — that b is also the length of the semi-minor axis, so the distance across the ellipse in the "shorter" direction is 2b. Yes, the Pythagorean Theorem is involved in proving this stuff. Yes, these are the same letters used in the Pythagorean Theorem. No, this is not the same as the Pythagorean Theorem. Yes, this is very confusing. Accept it, make sure to memorize the relationship before the next test, and move on.)

OK   then      b^2 +c^2 = a^2       we know b = 3    c=3 (didstance from center to one of the foci) so we have:

3^2 +3^2 = a^2       thus a^2 = 18    and  FINALLY, we have the quation of the ellipse in question:

(x-1)^2 / 3^2   +  (y-2)^2 / 18  =1

(there PROBABY is an easier way !!!!   This was my thought process as I was reading through Purplemath learning about ellipses....you might want to read through it too ! )

Here is a desmos graph:

ElectricPavlov Jan 21, 2018
edited by ElectricPavlov  Jan 21, 2018
edited by ElectricPavlov  Jan 21, 2018
edited by ElectricPavlov  Jan 21, 2018
#2
+100521
0

Nice, EP!!!

Jan 21, 2018