1. Find the equations of the tangents to the hyperbola 9x2 - 16y2 = 1 drawn parallel to 9x + 8y = 10.
2. Find the equations of tangents to the hyperbola x2-4y2 =4 drawn perpendicular to x +2y = 0.
4. Determine the equation to hyperbola whose centre at (0, 0), distance between the foci is 18 and distance between the directrices is 8.
Sol: Given C = (0, 0)
Distance between foci: 18
⇒ 2ae = 18
⇒ ae = 9 ......... (1)
Distance between Directrices = 8
⇒ 2a/e = 8
⇒ a/e = 4
multiplying (1) and (2)
⇒ (ae) a/e =36
⇒ a2 = 36
We know that b2 = a2 (e2 - 1)
⇒ b2 = 81 - 36
⇒ b2 = 45
6. Find the equation of the hyperbola whose centre is (1, 0), focus is (6, 0) and the length of transverse axis is 6.
Sol: Given: Centre: C (1, 0)
Focus: S (6, 0)
Length of transverse axis: 2a = 6
⇒ a = 3
We know that: CS = ae = | 1- 6 | = 5
⇒ ae = 5
Also, b2 = a2(e2 -1)
⇒ b2 = (ae)2 - a2
⇒ b2 = 25 - 9
⇒ b2 = 16
b = 4
Since it represents a hyperbola
then 9 - C > 0 and 5 - C < 0
⇒ 9 > C and 5 < C
⇒ C < 9 ............. (1) and 5 < C ............. (2)
from (1) and (2)
5 < C < 9
8. Find the equations of the normal to the hyperbola x2 -3y2 = 144 at the end of latus rectum in first quadrant.
a2 = 144; b2 = 48 ( ... a > b)
10. Find the equation of the hyperbola with eccentricity , focus (1, 2) and corresponding directrix is 2x + y = 1
Sol: Given : focus = S(1, 2)
Eccentricity = e =
Directrix = 2x + y- 1 = 0
Let P(x, y) be any point on the hyperbola
By definition of Hyperbola PS = ePM
⇒ 5(x2 + 1- 2x + y2 + 4 - 4y) = 3(4x2 + y2 + 1 + 4xy - 2y- 4x)
⇒ 12x2 + 3y2 + 3+12xy - 6y - 12x - 5x2 - 5 +10x - 5y2 - 20 + 20y = 0
⇒ 7x2 + 12xy − 2y2 - 2x + 14y - 22 = 0
11. Find the centre, eccentricity, foci, vertices, directrices and length of latus rectum of the hyperbola 9x2 - 16y2 + 72x - 32y - 16 = 0
Sol: Given hyperbola = 9x2 - 16y2 + 72x - 32y = 16 = 0
⇒ 9(x2 + 8x) − 16 (y2 + 2y) = 16
13. Show that the locus of poles w.r.t. the parabola y2 = 4ax of the tangents to the
rectangular hyperbola x2 - y2 = a2 lies on the ellipse 4x2 + y2 = 4a2
Sol: Given parabola: y2 = 4ax
Polar of (x1, y1) : yy1 = 2a(x + x1)
Given rectangular hyperbola : x2- y2 = a2
By data condition = c2 = a2m2 - b2
c2 = a2(m2-1) [ ... b = a]
⇒ 4x12 = 4a2 - y12
⇒ 4x12 + y12 = 4a2
Locus of P (x1, y1) : 4x2 + y2 = 4a2
14. If e, e1 are the eccentricities of a hyperbola and its conjugate hyperbola