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If the line $\mathrm{y}=\mathrm{x}$ is a tangent to the parabola $\mathrm{y}=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ at the point $(1,1)$ and the curve passes through $(-1,0)$, then
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Verified Answer
The correct answer is:
$a=c=\frac{1}{4}, b=\frac{1}{2}$
Hint:
$\frac{d y}{d x}=2 a x+b$
at $(1,1) \frac{d y}{d x}=2 a+b=1$ ...(1)
Now $(1,1)$ and $(-1,0)$ satisfies the curve
$a-b+c=0$ ...(2)
$a+b+c=1$ ...(3)
$\mathrm{b}=\frac{1}{2} ; \mathrm{a}=\frac{1}{4} ; \mathrm{c}=\frac{1}{4}$
$\frac{d y}{d x}=2 a x+b$
at $(1,1) \frac{d y}{d x}=2 a+b=1$ ...(1)
Now $(1,1)$ and $(-1,0)$ satisfies the curve
$a-b+c=0$ ...(2)
$a+b+c=1$ ...(3)
$\mathrm{b}=\frac{1}{2} ; \mathrm{a}=\frac{1}{4} ; \mathrm{c}=\frac{1}{4}$
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