Question

Equation of common tangent(s) of x^2-y^2 = 12 and xy = 8 is (are)???

Answer 1

equation of curves are ; x² - y² = 12 and xy = 8

we have to find common tangents.

let y = mx + c is the common tangent of given curves.

so, it satisfies both the curves.

x(mx + c) = 8

⇒mx² + cx - 8 = 0

Discriminant = c² + 4(-8)m = 0

or, c² - 32m = 0 ......(1)

again, x² - (mx + c)² = 12

⇒x² - m²x² - c² - 2mcx = 12

⇒(1 - m²)x² - 2mcx - c² - 12 = 0

so, discriminant = (-2mc)² -4{-(c² + 12)(1-m²)} = 0

or, 4m²c² + 4(c² + 12)(1 - m²) = 0

or, m²c² + c² - c²m² + 12 - 12m² = 0

or, c² + 12 - 12m² = 0......(2)

from equations (1) and (2),

32m + 12 - 12m² = 0

or, 12m² - 32m - 12 = 0

or, 3m² - 8m - 3 = 0

or, 3m² - 9m + m - 3 = 0

or, 3m(m - 3) + (m - 3) = 0

or, (3m + 1)(m - 3) = 0

or, m = 3, -1/3

and c² = 32m = 96, but c² ≠ -32/3

so, m ≠ -1/3

c² = 96 ⇒c = ±4√6

hence, equation of common tangents are y = 3x + 4√6 and y = 3x - 4√6 are common tangents of curves.