Question

y^2-8x+6y+9=0 find axis ,latus rectum ​

Answer 1

Answer:

ANSWER

y2−6y=−8x−1

y2−6y+9=−8x+1+9 (Adding 9 on both sides)

(y−3)2=−8x+8

(y−3)2=−8(x−1)

This is of the form Y2=−8X where X=x−1 and Y=y−3

4a=8⇒a=48=2 This type is open leftward

Referred to X,Y axes         Referred to x,y axes

where X=x−1 and Y=y−3

Axis                                                  

Y=0

Y=0⇒y−3=0

⇒y=3

Vertex

(0,0)X=0⇒x−1=0

⇒x=1

Y=0⇒y−3=0

⇒y=3

∴ Vertex is V(1,3)

Focus

(−a,0)

i.e., (−2,0)X=0⇒x−1=0

⇒x=1

Y=0⇒y−