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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

If the normal to the parabola $y^2 = 4ax$ intersects the axis of the parabola at a distance of $4a$ from the vertex,then the slopes of the normal are in which progression?

The ordinates of the vertices of a triangle inscribed in the parabola $y^2 = 4ax$ are $y_1, y_2, y_3$. Find the area of the triangle.

What is the focus of the parabola $4y^2 + 12x - 20y + 67 = 0$?

If the point $(a, 2a)$ is an interior point of the region bounded by the parabola $y^2 = 16x$ and the double ordinate through the focus,then

The tangent drawn at any point $P$ to the parabola ${y^2} = 4ax$ meets the directrix at the point $K$. Then the angle which $KP$ subtends at its focus is ............. $^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module