Answer
Verified
38.1k+ views
In a concave mirror if the ray of light is parallel to the axis of the mirror, then the reflected ray will pass through the focus of the concave mirror.
As per the information given, we can draw the diagram as follow.
Let the ray of light meet the concave mirror at point P. Then Pf is the reflected ray where f is the focus of the concave mirror.
The equation of parabola or the equation of the concave mirror is,
${{y}^{2}}=4ax$
So, the coordinate of the point f will be $\left( a,0 \right)$.
As it is given that ray of light with equation $y=b$ strikes the concave mirror, so substituting the value of ‘y’ we get,
${{b}^{2}}=4ax$
$x=\dfrac{{{b}^{2}}}{4a}$
So, the coordinates of the point P is $\left( \dfrac{{{b}^{2}}}{4a},b \right)$ .
Now the slope of the line passing through two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$is given by,
$m=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$
So, the slope of the line passing through point \[f\left( a,0 \right)\] and \[P\left( \dfrac{{{b}^{2}}}{4a},b \right)\] is,
$m=\dfrac{b-0}{\dfrac{{{b}^{2}}}{4a}-a}$
$\Rightarrow m=\dfrac{b}{\dfrac{{{b}^{2}}-4{{a}^{2}}}{4a}}$
$\Rightarrow m=\dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}}$
Now we know the line of equation passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$is,
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
where ‘m’ is the slope of the line.
So, the equation of line passing through the point \[f\left( a,0 \right)\]with slope, $m=\dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}}$, is,
$y-0=\left( \dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}} \right)\left( x-a \right)$
$\Rightarrow \left( {{b}^{2}}-4{{a}^{2}} \right)y=4ab\left( x-a \right)$
$\Rightarrow 4abx-4{{a}^{2}}b-\left( {{b}^{2}}-4{{a}^{2}} \right)y=0$
$\Rightarrow 4abx-\left( {{b}^{2}}-4{{a}^{2}} \right)y-4{{a}^{2}}b=0$
Now as the reflected ray is passing through the focus, i.e., f, so the equation of reflected ray is,
$4abx-\left( {{b}^{2}}-4{{a}^{2}} \right)y-4{{a}^{2}}b=0$
Note:- In this particular type of questions first find the intersection point of line and parabola, then find the slope of line then apply equation of line formula you will get your answer
As per the information given, we can draw the diagram as follow.
Let the ray of light meet the concave mirror at point P. Then Pf is the reflected ray where f is the focus of the concave mirror.
The equation of parabola or the equation of the concave mirror is,
${{y}^{2}}=4ax$
So, the coordinate of the point f will be $\left( a,0 \right)$.
As it is given that ray of light with equation $y=b$ strikes the concave mirror, so substituting the value of ‘y’ we get,
${{b}^{2}}=4ax$
$x=\dfrac{{{b}^{2}}}{4a}$
So, the coordinates of the point P is $\left( \dfrac{{{b}^{2}}}{4a},b \right)$ .
Now the slope of the line passing through two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$is given by,
$m=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$
So, the slope of the line passing through point \[f\left( a,0 \right)\] and \[P\left( \dfrac{{{b}^{2}}}{4a},b \right)\] is,
$m=\dfrac{b-0}{\dfrac{{{b}^{2}}}{4a}-a}$
$\Rightarrow m=\dfrac{b}{\dfrac{{{b}^{2}}-4{{a}^{2}}}{4a}}$
$\Rightarrow m=\dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}}$
Now we know the line of equation passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$is,
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
where ‘m’ is the slope of the line.
So, the equation of line passing through the point \[f\left( a,0 \right)\]with slope, $m=\dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}}$, is,
$y-0=\left( \dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}} \right)\left( x-a \right)$
$\Rightarrow \left( {{b}^{2}}-4{{a}^{2}} \right)y=4ab\left( x-a \right)$
$\Rightarrow 4abx-4{{a}^{2}}b-\left( {{b}^{2}}-4{{a}^{2}} \right)y=0$
$\Rightarrow 4abx-\left( {{b}^{2}}-4{{a}^{2}} \right)y-4{{a}^{2}}b=0$
Now as the reflected ray is passing through the focus, i.e., f, so the equation of reflected ray is,
$4abx-\left( {{b}^{2}}-4{{a}^{2}} \right)y-4{{a}^{2}}b=0$
Note:- In this particular type of questions first find the intersection point of line and parabola, then find the slope of line then apply equation of line formula you will get your answer
Recently Updated Pages
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
The number of ways in which 5 boys and 3 girls can-class-12-maths-JEE_Main
Find dfracddxleft left sin x rightlog x right A left class 12 maths JEE_Main
Distance of the point x1y1z1from the line fracx x2l class 12 maths JEE_Main
In a box containing 100 eggs 10 eggs are rotten What class 12 maths JEE_Main
dfracddxex + 3log x A ex cdot x2x + 3 B ex cdot xx class 12 maths JEE_Main
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
P Q and R long parallel straight wires in air carrying class 12 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
The reaction of Zinc with dilute and concentrated nitric class 12 chemistry JEE_Main
How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main