Question

Fine common tangent to hyperbola 9x^2 - 9y^2 = 8 and the parabola y^2 = 32x

9x+3y= 8

9x-3y= -8

9x-3y= -4

9x-3y= 8

Answer 1

Equation of tangent in terms of slope of y2 = 32x is

y = mx + 8/m …... (1)

which is also a tangent of 9x2 – 9y2 = 8

So, x2 – y2 = 8/9

Hence, (8/m)2 = 8/9 m2 – 8/9

So, 8/m2 = m2/9 – 1/9

so, 72 = m4 – m2

So, m4 – m2 – 72 = 0

(m2 – 9)(m2 + 8) = 0

So, m = 9 and m2 + 8 ≠ 0

So, m = ± 3

So, from eq (1), we have y = ± 3x ± 8/3

So, 3y = ± 9x ± 8

or ± 9x – 3y ± 8 = 0

i.e. 9x – 3y + 8 = 0, 9x – 3y – 8 = 0

-9x – 3y + 8 = 0, – 9x – 3y – 8 = 0

or 9x – 3y + 8 = 0, 9x – 3y – 8 = 0

9x + 3y - 8 = 0, 9x + 3y + 8 = 0