CircleHard

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

Create a free account to view solution

View Solution Free
Topic: Circle·Practice all Circle questions

More Circle Questions

The equation of the common tangent touching the circle (x -3)2 + y2 = 9 and the parabola y2 = 4x above the x-axis is...The circle x2 + y2 - 2 x - 3 k y - 2 = 0 passes through two fixed points, (k is the parameter)...From a point P tangent is drawn to the circle x2 + y2 = a2 and a tangent is drawn to x2 + y2 = b2. If these tangent are ...If the normal at point P to the rectangular hyperbola x2 - y2 = 4 meets the transverse and conjugate axes at A and B res...The equation(s) of the tangent at the point (0, 0) to the circle, making intercepts of length 2a and 2b units on the co-...