# Solved Examples - Hyperbola Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE

## JEE: Solved Examples - Hyperbola Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE

The document Solved Examples - Hyperbola Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
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Illustration 1: If e1 is the eccentricity of the ellipse x2/16 + y2/25 = 1 and e2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2 = 1, then find the equation of the hyperbola.(2006)
Solution: The eccentricity of x2/16 + y2/25 = 1 is
e= √(1-16/25) = 3/5
e2 = 5/3
This is obtained using the relation e1e2 = 1.
Hence, the foci of the ellipse are (0, ± 3)
Hence, the equation of the hyperbola is x2/16 - y2/9 = -1.

Illustration 2: A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x2 + 4y2 = 12. Then its equation is (2007)
1. x2cosec2 θ – y2sec2 θ = 1
2. x2sec2 θ – y2cosec2θ = 1
3. x2sin2 θ – y2cos2 θ = 1
4. x2cos2 θ – y2sin2 θ = 1
Solution: The given ellipse is x2/4 + y2/3 = 1
This gives a = 2 and b = √3
Hence, 3 = 4(1-e2) which gives e = 1/2
So, ae = 2.1/2 = 1
Hence, the eccentricity e1 for the hyperbola is given by
1 = esin θ which means e1 = cosec θ.
So, b2 = sin2 θ(cosec2 θ – 1) = cos2 θ
Hence, the equation of hyperbola is x2/sin2 θ - y2/cos2θ = 1
Or x2cosec2 θ – y2sec2 θ = 1

Illustration 3: The circle x2 + y2 – 8x = 0 and the hyperbola x2/9 - y2/4 = 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 + √9m2-4
Equation of tangent to circle is y = m(x-4) + √16m+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 x2/9 - y2/4 = 1 and that of circle is x2 + y– 8x = 0.
For their points of intersection x2/9 + (x2– 8x)/ 4 = 1
So, this gives 4x2 + 9x2 -72x = 36
So, 13x2 -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 (x-6)2 + (y+2√3)(y-2√3) = 0
This gives x+ y2 -12x + 24 = 0

The document Solved Examples - Hyperbola Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
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