Show that the locus of feet of the perpendiculars drawn from foci to any tangent of the hyperbola x2a2-y2b2 = 1 is the auxiliary circle of the hyperbola.

Let the equation of hyperbola be x2a2-y2b2 = 1Let p(x1, y1) be the foot of the perpendicular drawn from either of the foci to a tangent The equation of the tangent to the hyperbola bey=mx±a2m2−b2→(1)The equation of the perpendicular from either foci (±ae, o) on this tangent y=−1m(x±ae)→(2)As ‘p’ is the point of Intersetion of( 1) and (2) we havey1=mx1±a2m2−b2,y1=−1mx1±ae⇒y1−mx1=±a2m2−b2,my1+x1=±aeSquaring and adding⇒y1−mx12+my1+x12=a2m2−b2+a2e2⇒y12+m2x12−2x1y1m+m2y12+x12+2mx1y1=a2m2−a2e2−1+a2e2 ⇒x121+m2+y121+m2=a2m2−a2e2+a2+a2e2 ⇒1+m2⋅x12+y12=a21+m2⇒x12+y12=a2∴p lies on x2+y2=a2 which is a circle with centre at the orgin, the centre of hyperbola

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Show that the locus of feet of the perpendiculars drawn from foci to any tangent of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is the auxiliary circle of the hyperbola.

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Detailed Solution

Let the equation of hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Let p(x₁, y₁) be the foot of the perpendicular drawn from either of the foci to a tangent

The equation of the tangent to the hyperbola be $y = mx \pm \sqrt{a^{2} m^{2} - b^{2}} \to (1)$
The equation of the perpendicular from either foci ( $\pm$ ae, o) on this tangent $y = \frac{- 1}{m} (x \pm ae) \to (2)$
As ‘p’ is the point of Intersetion of( 1) and (2) we have
$\begin{array}{l} y_{1} = {mx}_{1} \pm \sqrt{a^{2} m^{2} - b^{2}}, y_{1} = \frac{- 1}{m} (x_{1} \pm ae) \\ \Rightarrow y_{1} - {mx}_{1} = \pm \sqrt{a^{2} m^{2} - b^{2}}, {my}_{1} + x_{1} = \pm ae \end{array}$
Squaring and adding
$\begin{array}{l} \Rightarrow {(y_{1} - {mx}_{1})}^{2} + {({my}_{1} + x_{1})}^{2} = a^{2} m^{2} - b^{2} + a^{2} e^{2} \\ \Rightarrow y_{1}^{2} + m^{2} x_{1}^{2} - 2 x_{1} y_{1} m + m^{2} y_{1}^{2} + x_{1}^{2} + 2 m x_{1} y_{1} = a^{2} m^{2} - a^{2} (e^{2} - 1) + a^{2} e^{2} \\ \Rightarrow x_{1}^{2} (1 + m^{2}) + y_{1}^{2} (1 + m^{2}) = a^{2} m^{2} - a^{2} e^{2} + a^{2} + a^{2} e^{2} \\ \Rightarrow (1 + m^{2}) \cdot (x_{1}^{2} + y_{1}^{2}) = a^{2} (1 + m^{2}) \Rightarrow x_{1}^{2} + y_{1}^{2} = a^{2} \end{array}$
$∴$ p lies on $x^{2} + y^{2} = a^{2}$ which is a circle with centre at the orgin, the centre of hyperbola