Question

The product of the perpendicular from two foci on any tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, is

• A. $a^2$
• B. $b^2$
• C. $-a^2$
• D. $-b^2$

Answer 1

Answer: (B), (D)

The correct options are
B $b^2$
D $-b^2$

Given: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Foci: $(\pm ae,0)$

Let the tangent be $y=mx+c$ where $c^2=m^2{a}^2-b^2$

Now, perpendicular on tangent drawn from $F$ and $F^{{}^{\prime }}$ at $M$ and $N$, Length of these perpendicular will be as

$FM=\frac{\mid 0-m\left(ae\right)-c\mid }{\sqrt{1^2+m^2}}$

$F^{{}^{\prime }}N=\frac{\mid 0-m(-ae)-c\mid }{\sqrt{1+m^2}}$

Now, their products, $FM\cdot F^{{}^{\prime }}N=\frac{\mid -(mae+c)\mid \cdot \mid (mae-c)\mid }{(1+m^2}$

Now, because of modulus there are posiibilities,

$FM\cdot F^{{}^{\prime }}N=\frac{\pm (mae+c)(mae-c)}{(1+m^2)}$

$=\frac{\pm (mae{)}^2-c^2}{1+m^2}=\frac{\pm (m^2{a}^2{e}^2-c^2)}{1+m^2}$

Putting the value of $c^2=\frac{\pm (m^2{a}^2{e}^2-a^2{m}^2+b^2)}{1+m^2}=\frac{\pm \left(m^2{a}^2\right(e^2-1)+b^2)}{1+m^2}$-----------1

In Equation 1, putting values of $b^2=a^2(e^2-1)$, we get

$FM\times F^{{}^{\prime }}N=\frac{\pm m^2{a}^2(e^2-1)+a^2(e^2-1)}{1+m^2}=\frac{\pm (m^2+1)\left(a^2\right(e^2-1\left)\right)}{1+m^2}$

$=\pm a^2(e^2-1)$

$FM\times F^{{}^{\prime }}N=\pm b^2$.