The equation of the tangent to the curve y^{2 | vedclass

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Question

(D) Given the curve equation: $y^{2} = ax^{2} + b$. Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = 2ax$ $\frac{dy}{dx} = \frac{ax

Vedclass · Accepted Answer

$6, -15$

Vedclass · Answer

(D) Given the curve equation: $y^{2} = ax^{2} + b$. Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = 2ax$ $\frac{dy}{dx} = \frac{ax}{y}$. The slope of the tangent at the point $(2, 3)$ is $\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{a(2)}{3} = \frac{2a}{3}$. The given equation of the tangent is $y = 4x - 5$,which is in the form $y = mx + c$,so the slope $m = 4$. Equating the slopes: $\frac{2a}{3} = 4 \Rightarrow 2a = 12 \Rightarrow a = 6$. Since the point $(2, 3)$ lies on the curve,it must satisfy the equation $y^{2} = ax^{2} + b$: $(3)^{2} = 6(2)^{2} + b$ $9 = 6(4) + b$ $9 = 24 + b$ $b = 9 - 24 = -15$. Thus,the values are $a = 6$ and $b = -15$.

The equation of the tangent to the curve $y^{2}=ax^{2}+b$ at the point $(2,3)$ is $y=4x-5$. Then the values of $a$ and $b$ are:

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