Question:
From the focus of the parabola y2=12x, a ray of light is directed in a direction making an angle \(tan^{-1}\frac{3}{4}\) with the x-axis. Then the equation of the line along which the reflected ray leaves the parabola is
From the focus of the parabola y2=12x, a ray of light is directed in a direction making an angle \(tan^{-1}\frac{3}{4}\) with the x-axis. Then the equation of the line along which the reflected ray leaves the parabola is
Updated On: Jun 26, 2024
- y=2
- y=18
- y=9
- y=36
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The Correct Option is B
Approach Solution - 1
The correct answer is option (B): y=18
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Approach Solution -2
Here, \(a = 3\)
So, the coordinates of the focus are (3, 0).
Let \(A\) be the focus of the parabola and \(AB\) be the incident ray and \(BC\) be the reflected ray.
Slope of \(AB = \tan\theta = \frac{3}{4}\)
Equation of \(AB\) is:
\(y - 0 = \frac{3}{4}(x - 3)\)
\(4y = 3x - 9\)
Let the coordinates of \(B\) be \(\left(\frac{a^2}{12}, a\right)\).
Now,
\(4\alpha = \frac{3\alpha^2}{12} - 9\)
\(16\alpha = \alpha^2 - 36\)
\(\alpha^2 - 16\alpha - 36 = 0\)
\(\alpha = 18\) [Neglecting \(\alpha = -2\)]
The reflected ray leaves the parabola at \(y = 18\).
So, the coordinates of the focus are (3, 0).
Let \(A\) be the focus of the parabola and \(AB\) be the incident ray and \(BC\) be the reflected ray.
Slope of \(AB = \tan\theta = \frac{3}{4}\)
Equation of \(AB\) is:
\(y - 0 = \frac{3}{4}(x - 3)\)
\(4y = 3x - 9\)
Let the coordinates of \(B\) be \(\left(\frac{a^2}{12}, a\right)\).
Now,
\(4\alpha = \frac{3\alpha^2}{12} - 9\)
\(16\alpha = \alpha^2 - 36\)
\(\alpha^2 - 16\alpha - 36 = 0\)
\(\alpha = 18\) [Neglecting \(\alpha = -2\)]
The reflected ray leaves the parabola at \(y = 18\).
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.