Question

The slopes of the common tangent to the hyperbolas

$\frac{x^{2}}{9}$– $\frac{y^{2}}{16}$= 1 and $\frac{y^{2}}{9}$–$\frac{x^{2}}{16}$= 1 are-

• A. –2
• B. –1
• C. 2
• D. None

Answer 1

Answer: (B)

–1

Answer 2

Let common tangent is y = mx + c ...(1)

Q (1) touch x²/9– y²/16 = 1

Ž C = ± $\sqrt{9m^{2} - 16}$

Q (1) also touch $\frac{x^{2}}{( - 16)}$ – $\frac{y^{2}}{( - 9)}$ = 1

Ž c = ± $\sqrt{- 16m^{2} + 9}$Ž 9m² – 16 = – 16m² + 9

Ž 25m² = 25 Ž m² = 1 Ž m = ± 1