# A ray of light is coming along the line $y=b$ from the positive direction of x-axis and strikes a concave mirror whose intersection with the \[xy\] -plane is a parabola ${{y}^{2}}=4ax$. Find the equation of the reflected ray and show that it passes through the focus of the parabola. Both a and b are positive.

Answer

Verified

383.1k+ 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.

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.

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

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is 1 divided by 0 class 8 maths CBSE

Find the HCF and LCM of 6 72 and 120 using the prime class 6 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

State the laws of reflection of light

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE

One card is drawn from a well shuffled deck of 52 playing class 12 maths CBSE