Question

answer me please please please
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Answer 1

Answer:

(a) $3x^{2}-y^{2}+10y-25$

Step-by-step explanation:

let the point where a line drawn ⊥ from P to the y axis : b and let pt (0,5)=A

let an arbitary pt P(x,y)

thus according to the question for any pt P, we have

AP=2BP [AP=distance of pt P from A & BP=distance of pt P from y axis{which           is P's x coordinate}]

now, AP=2BP

$\sqrt{(x-0)^{2}+(y-5)^{2} } =2x$

$x^{2} +(y-5)^{2}=(2x)^{2}$

$=4x^{2}$

⇒$3x^{2} -y^{2} +10y-25=0$ [transposing every thing to the other side]

=(a)

Answer 2

Answer:

Step-by-step explanation:

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BY Tolety Roshan