# ✓ show : x²/a² + y²/b² = 1 ➯ PF + PF' = 2a .?

F(c,0),F'(-c,0),P(x,y), b² + c² = a² or a > c.

### 2 Answers

- RealProLv 72 years agoBest Answer
Given point P(x,y) that satisfies equation x²/a² + y²/b² = 1

y² = b² - x²(b²/a²)

Distance from P(x,y) to F(c,0) is √[ (x-c)² + y² ]

Likewise distance from P(x,y) to F'(-c,0) is √[ (x+c)² + y² ]

PF = √[ (x-c)² + y² ] = √[ x² - 2xc + c² + b² - x²(b²/a²) ]

= √[ a² + x² - 2xc - x²(b²/a²) ]

= √[ a² - 2xc + x²(1 - b²/a²) ]

= √[ a² - 2xc + x²(a² - b²)/a² ]

= √[ a² - 2xc + x²c²/a² ]

= √ [ a - xc/a ]²

= a - xc/a

Similarly (just change sign of c)

PF' = √[ (x+c)² + y² ] = √[ x² + 2xc + c² + b² - x²(b²/a²) ]

= √[ a² + x² + 2xc - x²(b²/a²) ]

= √[ a² + 2xc + x²(1 - b²/a²) ]

= √[ a² + 2xc + x²(a² - b²)/a² ]

= √[ a² + 2xc + x²c²/a² ]

= √ [ a + xc/a ]²

= a + xc/a

PF + PF' = 2a

- Barry GLv 72 years ago
Suppose that P is on the x axis, where y=0. Then the x co-ordinate of P is $a$. The foci F. F' are at (0,c) and (0,-c) so PF=$a-c$ and PF'=$a+c$. Therefore PF+PF'=$2a$.