Question

The perpendicular from the focus on either asymptote meets it in the same points as the corresponding directrix & common points of intersection lie on the auxiliary circle.

• A. True
• B. False

Answer 1

The correct option is A

True

let's consider a general hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Equation of the asymptotes is $y=\pm \frac{b}{a}x$

Coordinates of the focus $(\pm ae,0)$

Let is the perpendicular from the foci (ae,0) to the asymptotes $y=\frac{b}{a}x$

slope of the $sp=\frac{1}{slopeofasymptotes}=\frac{-1}{\frac{b}{a}}=\frac{-a}{b}$

Equation of the sp is passes through (ae,0) with slope $\frac{-a}{b}$

$y-0=\frac{-a}{b}(x-ae)$

$y-=\frac{-a}{b}(x-ae)$ - - - - - - - - -(1)

Equation of asymptote is $y=\frac{a}{a}x$ - - - - - - - - -(2)

solving equation $\left(1\right)\&\left(2\right)$

we get $\frac{b}{a}x=\frac{-a}{b}(x-ae)$

$-\frac{b^2}{a^2}x=x-ae$

$-x-\frac{b^2}{a^2}x=x-ae$

$-x(1+\frac{b^2}{a^2})=-ae$

$Xxe2=ae$

$x=\frac{a}{e}$

point of intersection of perpendicular from focus to

the asymptote always lies on the directrix.

Here,$x=\frac{a}{e}$

substituting $x=\frac{a}{e}$ in the equation (2)

$y-\frac{b}{a}x=\frac{b}{a}\times \frac{a}{e}=\frac{b}{e}$

$y=\frac{b}{e}$

squaring and adding $x\&y$

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

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

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

so,we can say that the common point of intersection of

perpendicular from the focus to the asymptotes lies on the auxiliary circle.