If-P-x-y-F-1-3-0-F-2-3-0-and-16x-2-25y-2-400-then-PF-1-PF-2-

Question Number 279 by amandeep last updated on 25/Jan/15

If P (x, y), F_{1} = (3, 0), F_{2} = (- 3, 0) and

16 x^{2} + 25 y^{2} = 400 then {P F}_{1} + {P F}_{2} = ?

Answered by 123456 last updated on 18/Dec/14

{(4 x)}^{2} + {(5 y)}^{2} = 20^{2}

P (x, y) = (5 \cos θ, 4 \sin θ)

{(4 \cdot 5 \cos θ)}^{2} + {(5 \cdot 4 \sin θ)}^{2} = 20^{2}

P (x, y) = (5 \cos θ, 4 \sin θ)

F_{1} = (3, 0)

F_{2} = (- 3, 0)

{P F}_{1} = F_{1} - P = (3, 0) - (5 \cos θ, 4 \sin θ)

= (3 - 5 \cos θ, - 4 \sin θ)

{P F}_{1}^{2} = {(3 - 5 \cos θ)}^{2} + {(- 4 \sin θ)}^{2}

= 9 - 30 \cos θ + {25 \cos}^{2} θ + {16 \sin}^{2} θ

= 9 - 30 \cos θ + {9 \cos}^{2} θ + 16

= 25 - 30 \cos θ + {9 \cos}^{2} θ

= 5^{2} - 2 \cdot 5 \cdot 3 \cos θ + {(3 \cos θ)}^{2}

= {(5 - 3 \cos θ)}^{2}

{P F}_{1} = 5 - 3 \cos θ, 2 ⩽ 5 - 3 \cos θ ⩽ 8

{P F}_{2} = F_{2} - P = (- 3, 0) - (5 \cos θ, 4 \sin θ)

= (- 3 - 5 \cos θ, - 4 \sin θ)

{P F}_{2}^{2} = {(- 3 - 5 \cos θ)}^{2} + {(- 4 \sin θ)}^{2}

= 9 + 30 \cos θ + {25 \cos}^{2} θ + {16 \sin}^{2} θ

= 9 + 30 \cos θ + {9 \cos}^{2} θ + 16

= 25 + 30 \cos θ + {9 \cos}^{2} θ

= {(5 + 3 \cos θ)}^{2}

{P F}_{2} = 5 + 3 \cos θ, 2 ⩽ 5 + 3 \cos θ ⩽ 8

{P F}_{1} + {P F}_{2} = 5 - 3 \cos θ + 5 + 3 \cos θ = 10

Commented by 123456 last updated on 18/Dec/14

f ixed

thank you .

Commented by prakash jain last updated on 18/Dec/14

In step

{PF}_{1}^{2} = {(3 - 5 \cos θ)}^{2} + {(- 4 \sin θ)}^{2}

It is actually {PF}_{1}^{2} and not {PF}_{1}

Answered by prakash jain last updated on 18/Dec/14

y^{2} = (400 - 16 x^{2}) / 25

{PF}_{1}^{2} = {(x - 3)}^{2} + y^{2}

= x^{2} - 6 x + 9 + \frac{400 - 16 x^{2}}{25}

= \frac{25 x^{2} - 150 x + 225 + 400 - 16 x^{2}}{25}

= \frac{9 x^{2} - 150 x + 625}{25} = \frac{{(25 - 3 x)}^{2}}{5^{2}}

{PF}_{1} = \frac{25 - 3 x}{5} t a k i n g + v e v a l u e x ⩽ 5 ∵ 400 - 16 x^{2} = 25 y^{2} ⩾ 0

Similarly

{PF}_{2} = \frac{(3 x + 25)}{5}

{PF}_{1} + {PF}_{2} = \frac{50}{5} = 10