1.

Find the equations of the tangents drawn to the curve `y^2-2x^2-4y+8=0.` from point `(1,2)`

Answer» y-2=slope(x-1)
`4-2-8+8!=0`
Lets say y=mx+c
`(mx+c)^2-2x^2-4y+8=0`
`(m^2-2)x^2+(2mc-4m)x+(c^2-4c+8)=0`
`D=0=b^2-4ac`
`(2mc-4m)^2-4(m^2-2)(c^2-4c+8)=0`
`c=2pmsqrt(2m^2-4`
`y=mx+2pmsqrt(2m^2-4`
`1=m+2pmsqrt(2m^2-4`
`m=pm2`
`c=0,m=2`
`c=4,m=-2`
`y=2x`
`y=-2x+4`.


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