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Official Sr. Teacher Gr II NON-TSP MATHEMATICS (Held on :29 Oct 2018)

Option 1 : \(\rm \dfrac{x^2}{4}-\dfrac{y^2}{5}=1\)

Electric charges and coulomb's law (Basic)

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10 Questions
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__Concept:__

Let A = (x_{1},y_{1}) and B = (x_{2} ,y_{2}) be any two points.

Then **Distance between A And B** is given by distance formula.

AB = \(\rm \sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\)

**Calculation:**

Given: Difference of distance from points (3, 0) and (-3, 0) is 4

Consider A = (3, 0) and B = (-3, 0)

Let a point be p(x, y)

As we know distance is always positive

So, |PA - PB| = 4

\(\Rightarrow \rm \sqrt{(x-3)^2+(y-0)^2} - \sqrt{(x+3)^2+(y-0)^2}= 4 \\\Rightarrow \rm \sqrt{(x-3)^2+(y)^2}=4+\sqrt{(x+3)^2+(y)^2}\)

Squaring both sides, we get

\(\Rightarrow \rm (\sqrt{(x-3)^2+(y)^2})^2=(4+\sqrt{(x+3)^2+(y)^2})^2\\\rm \Rightarrow (x-3)^2+y^2=16+(x+3)^2+y^2+8\sqrt{(x+3)^2+(y)^2}\\\rm \Rightarrow x^2-6x+9 = 16+x^2+6x+9+8\sqrt{(x+3)^2+(y)^2}\\\rm \Rightarrow-12x-16=8\sqrt{(x+3)^2+(y)^2}\\\rm \Rightarrow-3x-4=2\sqrt{(x+3)^2+(y)^2}\)

Squaring both sides, we get

\(\rm \Rightarrow(-3x-4)^2=(2\sqrt{(x+3)^2+(y)^2})^2\\\rm \Rightarrow(3x+4)^2=4 ((x+3)^2+(y)^2)\\\rm \Rightarrow 9x^2+16+24x = 4(x^2+6x+9+y^2)\\\rm \Rightarrow 5x^2-4y^2 = 20\\\rm\therefore \frac{x^2}{4} - \frac{y^2}{5}=1\)