x

posted by .

If P(x,y),F1(3,0), F2(-3,0) a nd 16x^2+25y^2=400, then find PF1+PF2.

• x -

Hint: an ellipse is the locus of points whose distances from the two foci have a constant sum. That sum is the length of the major axis.

So, just write your equation in standard form for an ellipse, and you can read off the value of the semi-axes.

Similar Questions

1. math - probability

Imagine a LONG bike with a total of 4 wheels - 2 in the front and 2 in the back. For the bike to work at all, at least 1 front wheel and 1 back wheel must be operational. All 4 tires are the same and each, individually, has probability …
2. Algebra

There are two solutions to | 16x - 5 | = 3. The greatest solution is ___. Since the expression, 16x - 5, can be either positive or negative, solve for both. 16x - 5 = 3 16x = 8 x = .5 -(16x - 5) = 3 -16x + 5 = 3 -16x = -2 x = 1/8 You …
3. Math

how do you factor 16x^2-25y^2?
4. math

(6x-5y)^2= 36x-25y 6x-5y (4x+5y)^2 16x+25y 4x+5y Please check to see if right. If not corrected for me.
5. algebra 1

solve by substitution. 0.5x+0.25y=36 y+18=16x
6. math

Find the points where these conics intercept: 16x^2 - 5y^2 = 64 16x^2 + 25y^2 - 96x = 256
7. maths

If P(x,y),F1(3,0), F2(-3,0) a nd 16x^2+25y^2=400, then find PF1+PF2.
8. algebra 2

Find the points where these quadratic relations intersect. 16x^2-5y^2=64, 16x^2+25y^2-96x=256
9. Algebra 2

Standard form for both equation with steps 16x^2-96x+144+25y^2=400 16x^2+192x+576+25y^2=1600
10. Algebra 2

Standard form for both equation with steps 16x^2-96x+144+25y^2=400 16x^2+192x+576+25y^2=1600

More Similar Questions