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

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$$\Rightarrow 2401+9{ y }^{ 2 }-294y=49{ \left( x+2 \right) }^{ 2 }+49{ y }^{ 2 }+441-294y$$$$\Rightarrow 1960=49{ \left( x+2 \right) }^{ 2 }+40{ y }^{ 2 }$$$$\displaystyle\Rightarrow \frac { { \left( x+2 \right) }^{ 2 } }{ 40 } +\frac { { y }^{ 2 } }{ 49 } =1$$Maths

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