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Let the quadratic curve passing through the point (-1,0) and touching the line y=x at (1,1) be $y=f(x)$. Then the x-intercept of the normal to the curve at the point (\alpha, \alpha+1) in the first quadrant is ______________.

Option: 1

11


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & f(x)=(x+1)(a x+b) \\ & 1=2 a+2 b \\ & f^{\prime}(x)=(a x+b)+a(x+1) \\ & 1=(3 a+b) \\ & \Rightarrow b=1 / 4, a=1 / 4 \\ & f(x)=\frac{(x+1)^2}{4} \\ & f^{\prime}(\mathrm{x})=\frac{x}{2}+\frac{1}{2} \quad \alpha+1=\frac{(\alpha+1)^2}{4}, \alpha>-1 \\ & \alpha+1=4 \\ & \alpha=3 \end{aligned}

normal at (3, 4)

\begin{aligned} & y-4=-\frac{1}{2}(x-3) \\ & y=0 \quad x=8+3 \end{aligned}

Ans. 11

Posted by

rishi.raj

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