Question

Find the focal length of a point such that the sum of its distance from the points (0,2) and (0,-2) is equal to 6

Answer 1

Answer:

$⇒9h²+5k²=45$

Step-by-step explanation:

Let P(h, k) be any  point on the locus and the let A(0,2) and B(0,-2) be the given points.

$⇒\sqrt{(h-0)²+(k-2)²}+\sqrt{(h-0)²+(k+2)²}$

$⇒\sqrt{h²+(k-2)²}=6-\sqrt{h²+(k+2)²}$

$s.o.b.s$

$⇒h²+(k-2)²=36-\sqrt[12]{h²+(k+2)²}+h²+(k+2)²$

$⇒-8k-36=\sqrt[-12]{h²+(k+2)²}$

$⇒(2k+9)²=9(h²+(k+2)²)$

$⇒4k²+36k+81=9h²+9k²+36k+36$

$⇒9h²+5k²=45$

$Hence,$

locus of $(h,k)$ is $9x²+5y²=45$

Hope it helps :-)

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