The length of the latus rectum of the parabol | vedclass

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

Question

(C) Given equation: $4y^2 - 20y + 2x + 17 = 0$ Divide by $4$: $y^2 - 5y + \frac{1}{2}x + \frac{17}{4} = 0$ Complete the square for $y$: $(y - \frac{5}

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(C) Given equation: $4y^2 - 20y + 2x + 17 = 0$ Divide by $4$: $y^2 - 5y + \frac{1}{2}x + \frac{17}{4} = 0$ Complete the square for $y$: $(y - \frac{5}{2})^2 - \frac{25}{4} + \frac{1}{2}x + \frac{17}{4} = 0$ $(y - \frac{5}{2})^2 = -\frac{1}{2}x + 2$ $(y - \frac{5}{2})^2 = -\frac{1}{2}(x - 4)$ Comparing with $(y - k)^2 = -4a(x - h)$,we get $4a = \frac{1}{2} = 0.5$. Thus,the length of the latus rectum is $0.5$.

The length of the latus rectum of the parabola $4y^2 + 2x - 20y + 17 = 0$ is

Similar Questions

The equation of a straight line drawn through the focus of the parabola $y^2 = -4x$ at an angle of $120^\circ$ to the $x$-axis is:

What is the distance between the focus and the directrix of the parabola $x^2 = -8y$?

Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $y^{2} = -8x$.

If a normal to the parabola $y^2=12x$ at $A(3,-6)$ cuts the parabola again at $P$,then the equation of the tangent at $P$ is

If $x+y=k$ is a normal to the parabola $y^{2}=12x$,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module