A ray of light travelling along the line x = | vedclass

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

Question

(C) The equation of the parabola is $y^2 - 2y - 4x + 5 = 0$.Completing the square for $y$,we get $(y-1)^2 - 1 - 4x + 5 = 0$,which simplifies to $(y-1)

Vedclass · Accepted Answer

$y = -1$

Vedclass · Answer

(C) The equation of the parabola is $y^2 - 2y - 4x + 5 = 0$.Completing the square for $y$,we get $(y-1)^2 - 1 - 4x + 5 = 0$,which simplifies to $(y-1)^2 = 4x - 4$,or $(y-1)^2 = 4(x-1)$.This is a parabola with vertex $V(1, 1)$ and focus $F(1+1, 1) = (2, 1)$.The incident ray travels along the line $x = 2$,which is the line passing through the focus $F(2, 1)$ and parallel to the $y$-axis.According to the reflective property of a parabola,any ray of light parallel to the axis of the parabola and striking the mirror will pass through the focus after reflection,and conversely,any ray passing through the focus will be reflected parallel to the axis of the parabola.The axis of this parabola is the line $y = 1$.Since the incident ray passes through the focus $F(2, 1)$,the reflected ray must be parallel to the axis of the parabola,which is $y = 1$.Thus,the reflected ray is a horizontal line of the form $y = k$.From the provided image,the reflected rays are $y = 3$ and $y = -1$. Comparing this with the options,$y = -1$ is a valid reflected ray.

$A$ ray of light travelling along the line $x = 2$ strikes a parabolic mirror with the equation $y^2 - 2y - 4x + 5 = 0$ and gets reflected from its surface. Then,the equation of the reflected ray may be:

Similar Questions

For a parabola with focus $(2, 1)$ and directrix $2x - 3y + 1 = 0$,what is the equation of the latus rectum?

The points of contact $Q$ and $R$ of the tangents drawn from the point $P(2, 3)$ to the parabola $y^2 = 4x$ are

The equation of the common tangent to the parabolas $y^2 = 32x$ and $x^2 = 256y$ is

$A$ beam is supported at its ends by supports which are $12 \, m$ apart. Since the load is concentrated at its centre,there is a deflection of $3 \, cm$ at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection $1 \, cm$?

If the tangents at the endpoints $P$ and $Q$ of a chord of a parabola meet at point $T$,then the distances of points $P, T, Q$ from the focus of the parabola are in which progression?

Vedclass Test Series

Exam Paper Generator

Online Exam Module