How to solve this general second degree equation for vertex and length of latus rectum for a parabola - Maths - Conic Sections

Question

How to solve this general second degree equation for vertex and
length of latus rectum for a parabola

Lovina Kansal · Accepted Answer

Dear student
x2+y2-8x-24y-2xy+48=0⇒x-y2=8x+24y-48      
(1)Introducing an arbitraray constant k, we may write the equation (1) asx-y+k2=8x+24y-48+2kx-y+k2⇒x-y+k2=8x+24y-48+2kx-2ky+k2⇒x-y+k2=8+2kx+24-2ky+k2-48      
(2)Now, choose k,such that the linesx-y+k=0and 8+2kx+24-2ky+k2-48=0are at right angles
The condition for this is1×8+2k2k-24=-1⇒8+2k=-2k+24⇒4k=16⇒k=4For this value of k, the equation (2) becomes⇒x-y+42=8+2(4)x+24-2(4)y+42-48⇒(x-y+4)2=16x+16y-32⇒(x-y+4)2=16(x+y-2)i
e Y2=16XNow, Y=0 i
e x-y+4=0 is the axisand X=0 i,e x+y-2 =0 is the tangent at the vertexSolving X=0 and Y=0i
e x-y+4=0and x+y-2=0We get the coodinates of the vertex, which are -1,3The latus ractum is 16    ∵ if equation  of parabola is y2=4ax then latus ractum is 4a
Regards