The curve 3y^2 = 2ax^2 + 6b passes through th | vedclass

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(A) The equation of the given curve is $3y^2 = 2ax^2 + 6b$ . . . $(i)$Since the curve passes through the point $P(3, -1)$,we substitute $x = 3$ and $y

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$a = 1/2, b = -1$

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(A) The equation of the given curve is $3y^2 = 2ax^2 + 6b$ . . . $(i)$Since the curve passes through the point $P(3, -1)$,we substitute $x = 3$ and $y = -1$ into equation $(i)$:$3(-1)^2 = 2a(3)^2 + 6b$$3 = 18a + 6b$Dividing by $3$,we get $6a + 2b = 1$ . . . $(ii)$Now,we differentiate equation $(i)$ with respect to $x$ to find the gradient:$6y \frac{dy}{dx} = 4ax$Given that the gradient at $P(3, -1)$ is $-1$,we substitute $\frac{dy}{dx} = -1$,$x = 3$,and $y = -1$:$6(-1)(-1) = 4a(3)$$6 = 12a$$a = 1/2$Substituting $a = 1/2$ into equation $(ii)$:$6(1/2) + 2b = 1$$3 + 2b = 1$$2b = -2$$b = -1$Thus,the values are $a = 1/2$ and $b = -1$.

The curve $3y^2 = 2ax^2 + 6b$ passes through the point $P(3, -1)$ and the gradient of the curve at $P$ is $-1$. Then the values of $a$ and $b$ are:

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