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

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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