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Question

If a ray of light incident along the line x=5 gets reflected from the hyperbola x216y29=1 at point P, then

A
equation reflected line is 9x+40y+45=0
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B
equation reflected line is 9x40y+45=0
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C
area of PSS=22.5 sq. unit
(S,S are foci of hyperbola)
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D
area of PSS=11.25 sq. unit
(S,S are foci of hyperbola)
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Solution

The correct option is D area of PSS=11.25 sq. unit
(S,S are foci of hyperbola)
The given hyperbola is
x216y29=1
e=54
Its foci are (±5,0)
given line passes through (5,0)
Hence P(5,±94)
So the reflected ray must pass through (5,0) and P
Hence the required equation is
y=±9/4(5+5)(x+5)40y=±9(x+5)
​​​​Area of triangle bounded between the incident line, reflected line and transverse axis is
=12×2.25×10=11.25 sq. unit

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