Two parabolas x = cy^2 + ay + b and x = cy - | vedclass

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(C) Given the parabolas $x_1 = cy^2 + ay + b$ and $x_2 = cy - y^2$ touch at $(0, 1)$.Since $(0, 1)$ lies on both curves:For $x_2$: $0 = c(1) - (1)^2 \

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$-5$

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(C) Given the parabolas $x_1 = cy^2 + ay + b$ and $x_2 = cy - y^2$ touch at $(0, 1)$.Since $(0, 1)$ lies on both curves:For $x_2$: $0 = c(1) - (1)^2 \Rightarrow c = 1$.For $x_1$: $0 = c(1)^2 + a(1) + b$ $\Rightarrow 1 + a + b = 0$ $\Rightarrow a + b = -1$.Now,find the derivatives $\frac{dx}{dy}$ at $y = 1$:For $x_1$: $\frac{dx_1}{dy} = 2cy + a = 2(1)(1) + a = 2 + a$.For $x_2$: $\frac{dx_2}{dy} = c - 2y = 1 - 2(1) = -1$.Since the curves touch,their tangents at $(0, 1)$ are identical,so $\frac{dx_1}{dy} = \frac{dx_2}{dy}$ at $y = 1$.$2 + a = -1 \Rightarrow a = -3$.Using $a + b = -1$,we get $-3 + b = -1 \Rightarrow b = 2$.Finally,calculate $2a + b - c = 2(-3) + 2 - 1 = -6 + 2 - 1 = -5$.

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	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

Two parabolas $x = cy^2 + ay + b$ and $x = cy - y^2$ touch each other at the point $(0, 1)$. Then the value of $2a + b - c$ is:

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