If the line y=x is a tangent to the parabola | vedclass

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(C) The equation of the parabola is $y = ax^2 + bx + c$. Since the point $(1,1)$ lies on the parabola, we have $1 = a(1)^2 + b(1) + c$, which gives $a

Vedclass · Accepted Answer

$a=c=\frac{1}{4}, b=\frac{1}{2}$

Vedclass · Answer

(C) The equation of the parabola is $y = ax^2 + bx + c$. Since the point $(1,1)$ lies on the parabola, we have $1 = a(1)^2 + b(1) + c$, which gives $a + b + c = 1$ ...$(1)$. Since the point $(-1,0)$ lies on the parabola, we have $0 = a(-1)^2 + b(-1) + c$, which gives $a - b + c = 0$ ...$(2)$. Subtracting $(2)$ from $(1)$, we get $2b = 1$, so $b = \frac{1}{2}$. Substituting $b = \frac{1}{2}$ into $(1)$, we get $a + c = \frac{1}{2}$ ...$(3)$. The slope of the tangent to the parabola at any point $(x,y)$ is given by $\frac{dy}{dx} = 2ax + b$. At the point $(1,1)$, the slope of the tangent is $2a(1) + b = 2a + b$. Since the line $y=x$ is tangent at $(1,1)$, its slope is $1$. Thus, $2a + b = 1$. Substituting $b = \frac{1}{2}$, we get $2a + \frac{1}{2} = 1$, which implies $2a = \frac{1}{2}$, so $a = \frac{1}{4}$. From $(3)$, $c = \frac{1}{2} - a = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$. Thus, $a = \frac{1}{4}, b = \frac{1}{2}, c = \frac{1}{4}$.

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

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

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