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Question

Let an incident ray L1 = 0 gets reflected at point A(-2, 3) on hyperbola = 1 & passes through focus S(2, 0), then -

Options

A.equation of incident ray is x + 2 = 0
B.equation of reflected ray is 3x + 4y = 6
C.eccentricity, e = 2
D.length of latus rectum = 6

Solution

    
S ≡ (2, 0), S′ ≡ (-2, 0)
Using reflection property of hyperbola,
S′A is incident ray.
Equation of incident ray
S′A is x = -2
Equation of reflected ray
SP is 3x + 4y = 6.
Now 2ae = 4    ⇒   ae = 2        .... (i)
Point (-2, 3) lies on hyperbola,
∴  = 1    ⇒  = 1
on solving it we get    a = 4 (reject), a = 1     ..... (ii)
∴ Using (i) & (ii),    we get e = 2
length of latus rectum = 2a(e2 - 1) - 6

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