Question

What is the point of contact between the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Solving the equation of hyperbola and tangent will give the point of contact.

$\frac{x^2}{a^2}-\frac{(mx\pm \sqrt{a^2{m}^2-b^2}{)}^2}{b^2}=1$

$b^2{x}^2-a^2(m^2{x}^2+a^2{m}^2-b^2\pm 2mx\sqrt{a^2{m}^2-b^2}=a^2{b}^{2}0$

$x^2(b^2-a^2{m}^2)+(\pm 2m{a}^2\sqrt{a^2{m}^2-b^2}+a^4{b}^2)=0$

$\therefore x=\frac{\pm 2m{a}^2\sqrt{a^2{m}^2-b^2}\pm \sqrt{4m^2{a}^4(a^2{m}^2-b^2)-4(b^2-a^2{m}^2)(-a^4{m}^2)}}{2(b^2-a^2{m}^2)}$

$=\frac{\pm 2m{a}^2[\sqrt{a^2{m}^2-b^2}\pm \sqrt{a^2{m}^2-b^2+b^2-a^2{m}^2}]}{2(b^2-a^2{m}^2)}$

$=\frac{\pm a^{2}m\sqrt{a^2{m}^2-b^2}}{2(b^2-a^2{m}^2)}$

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

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

$\therefore$ The required point of contact is $(\frac{\pm a^{2}m}{\sqrt{a^2{m}^2-b^2}},\frac{\pm b^2}{\sqrt{a^2{m}^2-b^2}})$