The point on the curve 4y^2 - 4y + 2x - 1 = 0 | vedclass

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Question

(A) Given the curve equation is $4y^2 - 4y + 2x - 1 = 0$. To find the point where the tangent is parallel to the $Y$-axis,we need to find $\frac{dx}{d

Vedclass · Accepted Answer

$\left(1, \frac{1}{2}\right)$

Vedclass · Answer

(A) Given the curve equation is $4y^2 - 4y + 2x - 1 = 0$. To find the point where the tangent is parallel to the $Y$-axis,we need to find $\frac{dx}{dy}$ and set it to $0$,or find $\frac{dy}{dx}$ and set the denominator to $0$. Differentiating the equation with respect to $y$: $\frac{d}{dy}(4y^2 - 4y + 2x - 1) = \frac{d}{dy}(0)$ $8y - 4 + 2\frac{dx}{dy} = 0$ $2\frac{dx}{dy} = 4 - 8y$ $\frac{dx}{dy} = 2 - 4y$ For the tangent to be parallel to the $Y$-axis,$\frac{dx}{dy} = 0$. $2 - 4y = 0 \implies 4y = 2 \implies y = \frac{1}{2}$. Now,substitute $y = \frac{1}{2}$ into the original curve equation to find $x$: $4(\frac{1}{2})^2 - 4(\frac{1}{2}) + 2x - 1 = 0$ $4(\frac{1}{4}) - 2 + 2x - 1 = 0$ $1 - 2 + 2x - 1 = 0$ $2x - 2 = 0 \implies 2x = 2 \implies x = 1$. Thus,the point is $(1, \frac{1}{2})$.

The point on the curve $4y^2 - 4y + 2x - 1 = 0$ at which the tangent becomes parallel to the $Y$-axis is:

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