Step 1: Set up the moving point and its chord of contact.
Let the variable point be $P(x_1,y_1)$. For the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, the chord of contact from $P$ has equation $T=0$, that is \[ \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1 \]
Step 2: Build the pair of lines from the origin to the chord ends.
The two points where this chord meets the hyperbola, joined to the origin, give a pair of straight lines. We get them by making the hyperbola equation homogeneous of degree two using the chord line.
Step 3: Homogenize the hyperbola.
Multiply the right side $1$ of the hyperbola by the chord written as equal to $1$. This gives \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=\left(\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\right)^2 \] which is a second degree homogeneous equation in $x$ and $y$, representing the two lines from the origin.
Step 4: Collect the coefficients of $x^2$ and $y^2$.
Expanding, the coefficient of $x^2$ is $\dfrac{1}{a^2}-\dfrac{x_1^2}{a^4}$ and the coefficient of $y^2$ is $-\dfrac{1}{b^2}-\dfrac{y_1^2}{b^4}$.
Step 5: Apply the right angle condition.
Two lines through the origin given by $Ax^2+2Hxy+By^2=0$ are perpendicular when $A+B=0$. So \[ \left(\frac{1}{a^2}-\frac{x_1^2}{a^4}\right)+\left(-\frac{1}{b^2}-\frac{y_1^2}{b^4}\right)=0 \]
Step 6: Rearrange and replace by the running coordinates.
This gives $\dfrac{x_1^2}{a^4}+\dfrac{y_1^2}{b^4}=\dfrac{1}{a^2}-\dfrac{1}{b^2}$. Dropping the subscripts gives the locus, matching the option \[ \frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2} \]
\[ \boxed{\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}=\dfrac{1}{a^2}-\dfrac{1}{b^2}} \]