If the line y=4x-5 touches the curve y^{2}=ax | vedclass

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(C) The slope of the line $y=4x-5$ is $4$. Since the line touches the curve at $(2,3)$,the derivative of the curve at this point must equal the slope

Vedclass · Accepted Answer

$a=2, b=-7$

Vedclass · Answer

(C) The slope of the line $y=4x-5$ is $4$. Since the line touches the curve at $(2,3)$,the derivative of the curve at this point must equal the slope of the line. Given $y^{2}=ax^{3}+b$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 3ax^{2}$. Thus,$\frac{dy}{dx} = \frac{3ax^{2}}{2y}$. At the point $(2,3)$,the slope is $\frac{3a(2)^{2}}{2(3)} = \frac{12a}{6} = 2a$. Equating this to the slope of the line,we get $2a = 4$,which implies $a = 2$. Since the point $(2,3)$ lies on the curve $y^{2}=ax^{3}+b$,we substitute $x=2, y=3$ and $a=2$ into the equation: $3^{2} = 2(2)^{3} + b$ $9 = 2(8) + b$ $9 = 16 + b$ $b = 9 - 16 = -7$. Therefore,$a=2$ and $b=-7$.

If the line $y=4x-5$ touches the curve $y^{2}=ax^{3}+b$ at the point $(2,3)$,then

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