Finding the focus

Geometry Level 4

T h e f o c u s o f t h e p a r a b o l a 4 x 2 − 4 x y + y 2 − 8 x − 6 y + 5 = 0 i s The\quad focus\quad of\quad the\quad parabola\quad 4{ x }^{ 2 }-4xy+{ y }^{ 2 }-8x-6y+5=0\quad is

( 3 5 , 2 5 ) \left( \frac { 3 }{ 5 } ,\frac { 2 }{ 5 } \right) ( 2 5 , 3 5 ) \left( \frac { 2 }{ 5 } ,\frac { 3 }{ 5 } \right) ( 3 5 , 4 5 ) \left( \frac { 3 }{ 5 } ,\frac { 4 }{ 5 } \right) ( 4 5 , 3 5 ) \left( \frac { 4 }{ 5 } ,\frac { 3 }{ 5 } \right)

Samarth Sangam by samarth sangam , 23, India

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1 solution

Samarth Sangam
Sep 21, 2014

4 x 2 − 4 x y + y 2 − 8 x − 6 y + 5 = 0 t h e e q u a t i o n c a n b e r e w r i t t e n a s 4 ( x − 3 5 ) 2 − 4 ( x − 3 5 ) ( y − 1 5 ) + 4 x ( y − 1 5 ) + 8 y = − 1 P u t X = x − 3 5 , Y = y − 1 5 a n d s o l v i n g w e g e t f o c u s t o b e ( 4 5 , 3 5 ) 4{ x }^{ 2 }-4xy+{ y }^{ 2 }-8x-6y+5=0\\ the\quad equation\quad can\quad be\quad rewritten\quad as\\ 4{ \left( x-\frac { 3 }{ 5 } \right) }^{ 2 }-4\left( x-\frac { 3 }{ 5 } \right) \left( y-\frac { 1 }{ 5 } \right) +4x\left( y-\frac { 1 }{ 5 } \right) +8y=-1\\ Put\quad X=x-\frac { 3 }{ 5 } ,Y=y-\frac { 1 }{ 5 } \\ and\quad solving\quad we\quad get\quad focus\quad to\quad be\quad \left( \frac { 4 }{ 5 } ,\frac { 3 }{ 5 } \right)

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