Common slope to the hyperbola and its conjugate

Question

The slope of the common tangent to the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ & $\frac{y^2}{9}-\frac{x^2}{16}=1$ is

(A) -2

(B) -1

(C) 2

(D) None of these

My approach is as follow y=mx+c, For the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ we get intercept $c=\pm \sqrt{9m^2-16}$

For the hyperbola $\frac{y^2}{9}-\frac{x^2}{16}=1$ we get intercept $c=\pm \sqrt{16m^2-9}$

Comparing the y intercept "c" and squaring we get $9m^2-16=16m^2-9$

Which is equal to $m^2=-1$, Hence there is no possible slope. I want to verify whether my approach regarding the equation of tangent of conjugate hyperbola is correct

Answer 1

No common tangents

$$E_1: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1,~~~ E_c: \frac{x^2}{-a^2}+\frac{y^2}{b^2}=1$$ $$E_2: \frac{x^2}{-b^2}+\frac{y^2}{a^2}=1$$ In OP's question $a=3, b=4$. Note that in $E_2$ Please note that $E_1$ and $E_c$ are conjugate hperbolas. These two have two have two common asymptotes $y=\pm \frac{b^2}{a^2} x$ which may be called common tangents to them at infinity.

In OP"s question the two hyperbolas are not even conjugate hyperbolas. These two hyperbolas are intersecting hyperbolas. See the Fig. below. these two cannot have common tangents.

Common tagents on $E_1$ and $E_2$ with slope $m$ are: $$y=mx\pm \sqrt{9m^2-16}~~~(1) , ~~y=mx\pm \sqrt{9-16m^2}~~~(2($$ By equating their intercepts we get $m=\pm 1$. Finally. (1) and (2) give imaginary intercepts on $y$-axis and hence there is no common tangent on $E_1$ and $E_2$. See their intersection at four corners.