Find the equation of the hyperbola with vertices at 0 , ± 6 and eccentricity 5 3 .

Since the vertices of the hyperbola lie on the y - axis, the transverse axis is along the y - axis. The equation of the hyperbola is of the form, y 2 a 2 - x 2 b 2 = 1 . Here, a = length of semi-transverse axis, b = length of semi- conjugate axis. The vertices of such hyperbola are 0 , ± a and on being compared with the given coordinates, we get a = 6 . Also, its given that the eccentricity e = 5 3 . We know that, b 2 = a 2 e 2 - 1 , therefore, b 2 = 6 2 5 3 2 - 1 = 100 - 36 = 64 Substituting in the equation of the hyperbola, we get, y 2 36 - x 2 64 = 1

Find the equation of the hyperbola with vertices at 0 , ± 6 and eccentricity 5 3 .

Since the vertices of the hyperbola lie on the y - axis, the transverse axis is along the y - axis. The equation of the hyperbola is of the form, y 2 a 2 - x 2 b 2 = 1 . Here, a = length of semi-transverse axis, b = length of semi- conjugate axis. The vertices of such hyperbola are 0 , ± a and on being compared with the given coordinates, we get a = 6 . Also, its given that the eccentricity e = 5 3 . We know that, b 2 = a 2 e 2 - 1 , therefore, b 2 = 6 2 5 3 2 - 1 = 100 - 36 = 64 Substituting in the equation of the hyperbola, we get, y 2 36 - x 2 64 = 1

If the line y = m x + a 2 m 2 - b 2 touches the hyperbola x 2 a 2 - y 2 b 2 = 1 at the point a sec ϕ , b tan ϕ , show that ϕ = sin - 1 b a m

The slope of the tangent at any point on the hyperbola can be calculated by differentiating it w.r.t. x . 2 x a 2 - 2 y b 2 . d y d x = 0 ⇒ d y d x = b 2 x a 2 y ⇒ m = b 2 a sec Φ a 2 b tan Φ ⇒ m = b a sin Φ ⇒ sin Φ = b a m ⇒ Φ = sin - 1 b a m

If the chord through the points ( a sec θ , b tan θ ) and ( a sec ϕ , b tan ϕ ) on the hyperbola x 2 a 2 - y 2 b 2 = 1 passes through a focus, prove that, tan θ 2 tan ϕ 2 = 1 - e 1 + e .

To prove: tan θ 2 tan ϕ 2 = 1 - e 1 + e The equation of the chord joining ( a sec θ , b tan θ ) and ( a sec ϕ , b tan ϕ ) y - b tan θ = b tan θ - b tan ϕ a sec θ - a sec ϕ ( x - a sec θ ) ⇒ y - b tan θ = b a tan θ - tan ϕ sec θ - sec ϕ ( x - a sec θ ) ⇒ y - b sin θ cos θ = b a sin θ cos θ - sin ϕ cos ϕ 1 cos θ - 1 cos ϕ ( x - a cos θ ) ⇒ y cos θ - b sin θ cos θ = b a sin θ · cos ϕ - sin ϕ · cos θ cos ϕ - cos θ x cos θ - a cos θ ⇒ y cos θ - b sin θ = b a sin θ - ϕ - 2 sin ϕ + θ 2 sin ϕ - θ 2 x cos θ - a ⇒ y cos θ - b sin θ = b a 2 sin θ - ϕ 2 cos θ - ϕ 2 2 sin ϕ + θ 2 sin θ - ϕ 2 x cos θ - a ⇒ y cos θ - b sin θ = b a cos θ - ϕ 2 sin ϕ + θ 2 x cos θ - a Which is reduces to, x a cos θ - ϕ 2 - y b sin θ + ϕ 2 = cos θ + ϕ 2 If it passes through the focus ( a e , 0 ) , then a e a cos θ - ϕ 2 - 0 b sin θ + ϕ 2 = cos θ + ϕ 2 ⇒ e cos θ - ϕ 2 = cos θ + ϕ 2 ⇒ e = cos θ + ϕ 2 cos θ - ϕ 2 ⇒ 1 e = cos θ - ϕ 2 cos θ + ϕ 2 ⇒ 1 - e e + 1 = cos θ - ϕ 2 - cos θ + ϕ 2 cos θ - ϕ 2 + cos θ + ϕ 2 ⇒ 1 - e e + 1 = - 2 sin θ - ϕ 2 + θ + ϕ 2 2 sin θ - ϕ 2 - θ + ϕ 2 2 2 cos θ - ϕ 2 + θ + ϕ 2 2 cos θ - ϕ 2 - θ + ϕ 2 2 ⇒ 1 - e e + 1 = 2 sin θ 2 sin ϕ 2 2 cos θ 2 cos ϕ 2 ⇒ 1 - e e + 1 = tan θ 2 tan ϕ 2 Hence, proved.