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