A ray of light passing through (2, 1) is refl | vedclass

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

Question

(C) The image of the point $(2, 1)$ with respect to the $y$-axis is $(-2, 1)$.The reflected ray passes through $(-2, 1)$ and $(5, 3)$.The slope of the

Vedclass · Accepted Answer

$2x - 7y + 29 = 0$ or $2x - 7y - 7 = 0$

Vedclass · Answer

(C) The image of the point $(2, 1)$ with respect to the $y$-axis is $(-2, 1)$.The reflected ray passes through $(-2, 1)$ and $(5, 3)$.The slope of the reflected ray is $m = \frac{3 - 1}{5 - (-2)} = \frac{2}{7}$.The equation of the reflected ray is $y - 3 = \frac{2}{7}(x - 5)$,which simplifies to $2x - 7y + 11 = 0$.Let the equation of the other directrix be $2x - 7y + \lambda = 0$.The distance between the two directrices of an ellipse is $\frac{2a}{e}$.The distance of the focus from the directrix is $\frac{a}{e} - ae = \frac{a(1 - e^2)}{e} = \frac{8}{\sqrt{53}}$.Given $e = \frac{1}{3}$,we have $\frac{a(1 - 1/9)}{1/3} = 3a 	imes \frac{8}{9} = \frac{8a}{3} = \frac{8}{\sqrt{53}}$,so $a = \frac{3}{\sqrt{53}}$.The distance between the two directrices is $\frac{2a}{e} = 2 	imes \frac{3}{\sqrt{53}} 	imes 3 = \frac{18}{\sqrt{53}}$.The distance between the parallel lines $2x - 7y + 11 = 0$ and $2x - 7y + \lambda = 0$ is $\frac{|\lambda - 11|}{\sqrt{2^2 + (-7)^2}} = \frac{|\lambda - 11|}{\sqrt{53}}$.Equating the distances: $\frac{|\lambda - 11|}{\sqrt{53}} = \frac{18}{\sqrt{53}}$,so $|\lambda - 11| = 18$.This gives $\lambda - 11 = 18$ or $\lambda - 11 = -18$,so $\lambda = 29$ or $\lambda = -7$.The equations are $2x - 7y + 29 = 0$ or $2x - 7y - 7 = 0$.

Similar Questions

If the tangents on the ellipse $4x^2 + y^2 = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other,then $a^2$ is equal to

If the straight line $8x + 3\sqrt{2}y = 36$ touches the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 2$ at $(a, b)$,then $a + \sqrt{2}b =$

For the ellipse $4x^2 + y^2 - 8x + 2y + 1 = 0$,the focus and the equation of the directrix are respectively

The values of $c$ such that the line $y=4x+c$ touches the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ are

The eccentric angle of a point on the ellipse $x^2+3y^2=6$ lying at a distance of $2$ units from its centre is

Vedclass Test Series

Exam Paper Generator

Online Exam Module