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Question: The focus of the parabola \(4 y ^ { 2 } - 6 x - 4 y = 5\)is...

The focus of the parabola 4y2−6x−4y=54 y ^ { 2 } - 6 x - 4 y = 5is

A

(−8/5,2)( - 8 / 5,2 )

B

(−5/8,1/2)( - 5 / 8,1 / 2 )

C

(1/2,5/8)( 1 / 2,5 / 8 )

D

(6/8,−1/2)( 6 / 8 , - 1 / 2 )

Answer

(−5/8,1/2)( - 5 / 8,1 / 2 )

Explanation

Solution

Given equation of parabola when written in standard form, we get

4(y−12)2=6(x+1)4 \left( y - \frac { 1 } { 2 } \right) ^ { 2 } = 6 ( x + 1 )⇒ (y−12)2=32(x+1)\left( y - \frac { 1 } { 2 } \right) ^ { 2 } = \frac { 3 } { 2 } ( x + 1 ) ⇒ Y2=32XY ^ { 2 } = \frac { 3 } { 2 } X

where, Y=y−12,X=x+1Y = y - \frac { 1 } { 2 } , X = x + 1

∴y=Y+12,x=X−1y = Y + \frac { 1 } { 2 } , x = X - 1 ....(i)

Focus ⇒ X=a,Y=0X = a , \quad Y = 0 ;

∴4a=324 a = \frac { 3 } { 2 } ⇒ a=38a = \frac { 3 } { 8 }

⇒ x=38−1=−58x = \frac { 3 } { 8 } - 1 = - \frac { 5 } { 8 } ; y=0+12=12y = 0 + \frac { 1 } { 2 } = \frac { 1 } { 2 } (−58,12)\left( - \frac { 5 } { 8 } , \frac { 1 } { 2 } \right)