If the lines x+y=a and x-y=b touch the curve | vedclass

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(D) The curve is given by $y = x^{2}-3x+2$.To find the points where the curve intersects the $x$-axis,we set $y = 0$:$x^{2}-3x+2 = 0$$(x-1)(x-2) = 0$S

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$0.50$

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(D) The curve is given by $y = x^{2}-3x+2$.To find the points where the curve intersects the $x$-axis,we set $y = 0$:$x^{2}-3x+2 = 0$$(x-1)(x-2) = 0$So,the points of intersection are $A(1, 0)$ and $B(2, 0)$.The slope of the tangent to the curve is given by $\frac{dy}{dx} = 2x - 3$.At point $A(1, 0)$,the slope is $m_1 = 2(1) - 3 = -1$.At point $B(2, 0)$,the slope is $m_2 = 2(2) - 3 = 1$.The line $x+y=a$ has a slope of $-1$. Since it touches the curve at $A(1, 0)$,we substitute the coordinates into the line equation:$1 + 0 = a \implies a = 1$.The line $x-y=b$ has a slope of $1$. Since it touches the curve at $B(2, 0)$,we substitute the coordinates into the line equation:$2 - 0 = b \implies b = 2$.Therefore,$\frac{a}{b} = \frac{1}{2} = 0.50$.

If the lines $x+y=a$ and $x-y=b$ touch the curve $y = x^{2}-3x+2$ at the points where the curve intersects the $x$-axis,then $\frac{a}{b}$ is equal to:

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