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Question

The perimeter of a triangle is $$20$$ and the points $$(-2,-3)$$ and $$(-2,3)$$ are two of the vertices of it. Then the locus of third vertex is:


A
(x2)249+y240=1
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B
(x+2)249+y240=1
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C
(x+2)240+y249=1
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D
(x2)240+y249=1
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Solution

The correct option is B $$\displaystyle \frac{(x+2)^{2}}{40}+\frac{y^{2}}{49}=1$$
Let the coordinates of the third vertex be $$A:\left( x,y \right) $$
Given that perimeter is $$20$$
The other two vertices given are $$B:\left( -2,-3 \right) $$ and $$C:\left( -2,3 \right) $$
We know that Perimeter $$=AB+AC+BC$$  ........(1)
Using distance formula, we get
$$\Rightarrow AB=\sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y+3 \right)  }^{ 2 } } $$
$$\Rightarrow AC=\sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } } $$
$$\Rightarrow BC=\sqrt { 36 } =6$$
Putting these values in (1), we get 
$$\sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y+3 \right)  }^{ 2 } } +\sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } } =20-6=14$$
$$\Rightarrow \sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y+3 \right)  }^{ 2 } } =14-\sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } } $$
Squaring both sides, we get
$${ \Rightarrow \left( x+2 \right)  }^{ 2 }+{ \left( y+3 \right)  }^{ 2 }=196+{ \left( x+2 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 }-28\sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } } $$
$$\Rightarrow 12y-196=-28\sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } } $$
$$\Rightarrow 49-3y=7\sqrt { { \left( x+2 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } } $$
Again squaring both sides, we get
$$\Rightarrow 2401+9{ y }^{ 2 }-294y=49{ \left( x+2 \right)  }^{ 2 }+49{ y }^{ 2 }+441-294y$$
$$\Rightarrow 1960=49{ \left( x+2 \right)  }^{ 2 }+40{ y }^{ 2 }$$
$$\displaystyle\Rightarrow \frac { { \left( x+2 \right)  }^{ 2 } }{ 40 } +\frac { { y }^{ 2 } }{ 49 } =1$$

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