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If y=mx+6is a tangent to both the parabolas y^2=8xand x^2=3by , thenbis equal to : |
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Answer» 36 So, `6=2/m rArr m=1/3` Now `y=x/3+6` is tangent to `x^2=3by` `rArr x^2=(3b)/(x/3+6)` `x^2=bx+18b` `x^2-bx-18b=0` D=0 `B^2=4 XX (-18) b^2 (b NE 0)` b=-72 |
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