Find the length of the latus rectum of the pa | vedclass

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

Question

(A) The given equation is $y = x 	an \alpha - \frac{g x^2}{2 u^2 \cos^2 \alpha}$.Rearranging the terms to the standard form of a parabola $x^2 = -4ay

Vedclass · Accepted Answer

$\frac{2 u^2 \cos^2 \alpha}{g}$

Vedclass · Answer

(A) The given equation is $y = x 	an \alpha - \frac{g x^2}{2 u^2 \cos^2 \alpha}$.Rearranging the terms to the standard form of a parabola $x^2 = -4ay$:$\frac{g x^2}{2 u^2 \cos^2 \alpha} = x 	an \alpha - y$$x^2 = \frac{2 u^2 \cos^2 \alpha}{g} (x 	an \alpha - y)$.This is of the form $(x - h)^2 = -4a(y - k)$.Comparing the coefficient of $y$,we have $4a = \frac{2 u^2 \cos^2 \alpha}{g}$.The length of the latus rectum is $4a = \frac{2 u^2 \cos^2 \alpha}{g}$.

Find the length of the latus rectum of the parabola $y = x \tan \alpha - \frac{g x^2}{2 u^2 \cos^2 \alpha}$.

Similar Questions

The equation of the directrix of the parabola $(2 x - 3 y - 5)^2 = 20(3 x + 2 y + 1)$ is

The distance of the point $(6, 4 \sqrt{3})$ from the focus of the parabola $y^2 = 8x$ is:

If $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a focal chord of the parabola $y^2 = 4ax$,then what is the square of the $G.M.$ of $x_1$ and $x_2$?

If the line segment joining the vertex of the parabola $y^2=4ax$ and a point on the parabola makes an angle $\theta$ with the positive $X$-axis,then the length of that line segment is

The length of the chord of the parabola $x^2 = 4ay$ passing through its vertex and having slope $\tan\alpha$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module