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Question Number 171315 by pete last updated on 12/Jun/22

$$\mathrm{The}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{curve}\mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+2$$$$\mathrm{at}(1,\frac{1}{2})\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{normal}\mathrm{to}\mathrm{the}\mathrm{curve}$$$$\mathrm{y}={\mathrm{x}}^2+6\mathrm{x}+10\mathrm{at}(-2,2).\mathrm{Find}\mathrm{the}\mathrm{values}$$$$\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}.$$

Answered by som(math1967) last updated on 12/Jun/22

$$Tangenttothecurvey={ax}^2+bx+2$$$$at(1,\frac{1}{2})\therefore \frac{1}{2}=a+b+2$$$$\Rightarrow a+b=-\frac{3}{2}.....i)$$$$\Rightarrow$$$$slopeoftangent$$$$\frac{dy}{dx}=2ax+b$$$${\left[\frac{dy}{dx}\right]}_{1,\frac{1}{2}}=2a+b$$$$slopeofnormal$$$$y=x^2+6x+10$$$$-\frac{dx}{dy}=\frac{-1}{2\times -2+6}=\frac{-1}{2}$$$$\therefore 2a+b=-\frac{1}{2}......ii)$$$$fromi)andii)$$$$a=-\frac{1}{2}+\frac{3}{2}=1$$$$b=-\frac{3}{2}-1=\frac{-5}{2}$$

Commented by pete last updated on 12/Jun/22

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir},\mathrm{i}\mathrm{am}\mathrm{grateful}.$$

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