WBJEE · Maths · Parabola
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
- A \(a=b=-1, c=3\)
- B \(a=b=\frac{1}{2}, c=0\)
- C \(a=c=\frac{1}{4}, b=\frac{1}{2}\)
- D \(a=0, b=c=\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(a=c=\frac{1}{4}, b=\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
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) \(\mathrm{b}=\frac{1}{2} ; \mathrm{a}=\frac{1}{4} ; \mathrm{c}=\frac{1}{4}\)
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