Question

Locus of the feet of the perpendicular drawn from focus of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ upon any tangent is _____

• A. x² + y² = 2a²
• B. x² + y² = a²
• C. x² - y² = a²
• D. x² - y² = 2a²

Answer 1

The correct option is B

x² + y² = a²

Equation of the given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Let slope of the tangent be m

Equation of the tangent to the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$y=mx\pm \sqrt{a^2{m}^2-b^2}$

Let coordinates of foot of perpendicular on tangent be (h,k) substituting (h,k) in the equation of

tangentg and squaring both sides

$(k-mh{)}^2=a^2{m}^2-b^2$ - - - - - - (1)

Slope of the foot of perpendicular is $-\frac{1}{slopeoftangent}$

$=-\frac{1}{m}$

Equation of feet of perpendicular drawn from the focus is $s(\pm ae,0)$

$y-0=\frac{-1}{m}(x\pm ae)$ - - -- - - - - (2)

$y=\frac{-1}{m}(x\pm ae)$

Equation (2) also passes through (h,k)

$k=\frac{-1}{m}(h\pm ae)$

$km+h=\pm ae$

Squaring on both sides

$(km+h{)}^2=a^2{e}^2=a^2+b^2=a^2+b^2$ - - - -- - - (3)

$\{e^2=1+\frac{b^2}{a^2}\}$

Adding equation (1) & (3)

so that we can eliminate (m)

$(k-mh{)}^2+(km+h{)}^2=a^2{m}^2-b^2+a^2+b^2$

$h^2+k^2+m^2(h^2+k^2)=a^2(1+m^2)$

$(h^2+k^2)(1+m^2)=a^2(1+m^2)$

$h^2+k^2-a^2$

For required locus $x^2+y^2=a^2$

so the locus the feet of the perpendicular drawn from locus of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ upon any

tangent is $x^2+y^2=a^2$ i.e.,auxiliary circle.

option B is correct.