Illustration 1: If e_{1} is the eccentricity of the ellipse x^{2}/16 + y^{2}/25 = 1 and e_{2} is the eccentricity of the hyperbola passing through the foci of the ellipse and e_{1}e_{2} = 1, then find the equation of the hyperbola.(2006)
Solution: The eccentricity of x^{2}/16 + y^{2}/25 = 1 is
e_{1 }= √(116/25) = 3/5
e_{2} = 5/3
This is obtained using the relation e_{1}e_{2} = 1.
Hence, the foci of the ellipse are (0, ± 3)
Hence, the equation of the hyperbola is x^{2}/16  y^{2}/9 = 1.
Illustration 2: A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x^{2} + 4y^{2} = 12. Then its equation is (2007)
1. x^{2}cosec^{2} θ – y^{2}sec^{2} θ = 1
2. x^{2}sec^{2} θ – y^{2}cosec^{2}θ = 1
3. x^{2}sin^{2} θ – y^{2}cos^{2} θ = 1
4. x^{2}cos^{2} θ – y^{2}sin^{2} θ = 1
Solution: The given ellipse is x^{2}/4 + y2/3 = 1
This gives a = 2 and b = √3
Hence, 3 = 4(1e^{2}) which gives e = 1/2
So, ae = 2.1/2 = 1
Hence, the eccentricity e_{1} for the hyperbola is given by
1 = e_{1 }sin θ which means e_{1} = cosec θ.
So, b^{2} = sin^{2} θ(cosec^{2} θ – 1) = cos^{2} θ
Hence, the equation of hyperbola is x^{2}/sin^{2} θ  y^{2}/cos^{2}θ = 1
Or x^{2}cosec^{2} θ – y^{2}sec^{2} θ = 1
Illustration 3: The circle x^{2} + y^{2} – 8x = 0 and the hyperbola x^{2}/9  y^{2}/4 = 1 intersect at the points A and B. (2010)
1. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is ….?
2. Also find the equation of the circle with AB as its diameter.
Solution: (1) Equation of tangent to hyperbola having slope m is y = mx + √9m^{2}4
Equation of tangent to circle is y = m(x4) + √16m^{2 }+16
These two equations will be identical for m = 2/√5
Hence, the equation of common tangent is 2x  √5y + 4 = 0
(2) The equation of the hyperbola is x^{2}/9  y^{2}/4 = 1 and that of circle is x^{2} + y^{2 }– 8x = 0.
For their points of intersection x^{2}/9 + (x^{2}– 8x)/ 4 = 1
So, this gives 4x^{2} + 9x^{2} 72x = 36
So, 13x^{2} 72x 36 = 0
This gives x = 6 and 13/6
But x = 13/6 is not acceptable
Now, x = 6, y = ± 2√3
Required equation is (x6)^{2} + (y+2√3)(y2√3) = 0
This gives x^{2 }+ y^{2} 12x + 24 = 0
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