Answer
64.8k+ views
Hint: As the reflected ray passes through the focus and we know the coordinates of the focus. So first we will find the intersection of a ray of light and a concave mirror and then use the equation of line formula to find the required equation.
Complete step-by-step answer:
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 follows.
![](https://www.vedantu.com/question-sets/c92c6b83-81f2-486e-89c6-04f27d5a1d381563439555170871488.png)
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 instead using the slope form of the equation of line we can use the point form of equation of line formula as well. In both cases you will get the same answer.
Complete step-by-step answer:
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 follows.
![](https://www.vedantu.com/question-sets/c92c6b83-81f2-486e-89c6-04f27d5a1d381563439555170871488.png)
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 instead using the slope form of the equation of line we can use the point form of equation of line formula as well. In both cases you will get the same answer.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)