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The slope of the tangent to the curve `x=t^2+3t-8,``y=2t^2-2t-5`at the point `(2, 1)`is(A) `(22)/7` (B) `6/7` (C) `7/6` (D) `(-6)/7`A. `(22)/(7)`B. `(6)/(7)`C. -6D. none of these |
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Answer» Correct Answer - B For the point (2, -1) on the curve `x=t^(2)+3t-8,y=2t^(2)- 2t -5` we have `t^(2)+3t-8=2 and 2t^(2)-2t-5=-1` `rArr t^(2)+3t-10=0 and 2t^(2)-2t-4=0` `rArr (t+5)(t-2)=0 and (t-2)(t+1)=0` `rArr t=2` Now, `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))=(4t-2)/(2t+3) rArr ((dy)/dx)_(t=2)=(4xx2-2)/(2xx2+3)=(6)/(7)` |
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