Question

Combined Equation of L and l1 is 2x²+axy+3y² and combined equatio. of l and l2 is 2x²+bxy-3y².If l1 is perpendicular to l2 then a²+b² is
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Answer 1

Answer: 26

Step-by-step explanation:

Let L⇒y=mxL⇒y=mx,

L1⇒y=kxL1⇒y=kx

L2⇒y=−1kxL2⇒y=−1kx

Now, (y−mx)(y−kx)=0(y−mx)(y−kx)=0

y2−ykx−mxy+mkx2=0y2−ykx−mxy+mkx2=0

⇒mkx2−(k+m)xy+y2=0⇒mkx2−(k+m)xy+y2=0

Given that, 2x2+axy+3y2=02x2+axy+3y2=0

or 23x2+a3xy+y2=023x2+a3xy+y2=0

On comparing, mk=23′−(k+m)=a3…mk=23′−(k+m)=a3…(i)

Now, (y−mx)(y+xk)=0(y−mx)(y+xk)=0

⇒y2+xyk−mxy−mx2k=0⇒y2+xyk−mxy−mx2k=0

⇒−mkx2+(1k−m)xy+y2=0⇒−mkx2+(1k−m)xy+y2=0

which is 2x2+bxy−3y2=02x2+bxy−3y2=0

or −23x2−b3xy+y2=0−23x2−b3xy+y2=0

On comparing, −mk=−23,−b3=1k−m−mk=−23,−b3=1k−m

So, m=2k3,−b3=1−kmk…m=2k3,−b3=1−kmk…(ii)

By solving Eqs. (i) and (ii), we get

m=23,k=1m=23,k=1 and a=−5,b=−1a=−5,b=−1

∴a2+b2=25+1=26