Question

The line $y=x$ is a tangent to the parabola $y=a{x}^2+bx+c$ at the point $x=1$.If the parabola passes through the point $(-1,0)$, then determine $a,b,c.$

• A. $a=\frac{1}{2},b=\frac{1}{4},c=\frac{1}{3}.$
• B. $a=\frac{1}{4},b=\frac{1}{2},c=\frac{1}{4}.$
• C. $a=2,b=1,c=4.$
• D. $a=4,b=2,c=4.$

Answer 1

The correct option is B $a=\frac{1}{4},b=\frac{1}{2},c=\frac{1}{4}.$

Given equation of parabola is
$y=a{x}^2+bx+c$
$\frac{dy}{dx}=2ax+b$
Slope of tangent to the curve at $x=1$ is $2a+b$
Given tangent is $y=x$ . Slope of this tangent is $1.$
So, $2a+b=1$ ...(1)
Since, the parabola passes through $(-1,0)$
$\therefore a-b+c==0$ ...(2)
Given $y=x$ is a tangent at $x=1$
$\therefore y=1.$
Hence $(1,1)$ lies both on tangent and parabola
$\therefore a+b+c=1$ ...(3)
Solving (1), (2) and (3), we get
$a=\frac{1}{4},b=\frac{1}{2},c=\frac{1}{4}.$