The locus of the point of intersection of any | vedclass

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(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Any tangent to the hyperbola is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$

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$x^2 + y^2 = a^2 - b^2$

Vedclass · Answer

(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Any tangent to the hyperbola is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.$A$ tangent perpendicular to this one will have a slope of $-\frac{1}{m}$,so its equation is $y = -\frac{1}{m}x \pm \sqrt{a^2(-\frac{1}{m})^2 - b^2} = -\frac{1}{m}x \pm \sqrt{\frac{a^2}{m^2} - b^2}$.Rearranging the equations: $(y - mx)^2 = a^2m^2 - b^2$ and $(my + x)^2 = a^2 - b^2m^2$.Adding these two equations: $(y - mx)^2 + (my + x)^2 = a^2m^2 - b^2 + a^2 - b^2m^2$.$y^2 - 2mxy + m^2x^2 + m^2y^2 + 2mxy + x^2 = a^2(1 + m^2) - b^2(1 + m^2)$.$(x^2 + y^2)(1 + m^2) = (a^2 - b^2)(1 + m^2)$.Since $1 + m^2 
eq 0$,we get $x^2 + y^2 = a^2 - b^2$.

The locus of the point of intersection of any two perpendicular tangents to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is a circle,which is called the director circle of the hyperbola. What is the equation of this circle?

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