Question

What is the point of contact between the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and
the tangent $y=mx\pm \sqrt{a^2{m}^2-b^2}$.

• A. $[\pm \frac{a^{2}m}{\sqrt{a^2{m}^2-b^2}},\pm \frac{b^2}{\sqrt{a^2{m}^2-b^2}}]$
• B. $[\pm \frac{b^2}{\sqrt{a^2{m}^2-b^2}},\pm \frac{a^{2}m}{\sqrt{a^2{m}^2-b^2}}]$
• C. $[\pm \sqrt{a^2{m}^2-b^2},\pm \sqrt{a^2{m}^2+b^2}]$
• D. $[\pm a^{2}m,\pm b^2]$

Answer 1

The correct option is A

$[\pm \frac{a^{2}m}{\sqrt{a^2{m}^2-b^2}},\pm \frac{b^2}{\sqrt{a^2{m}^2-b^2}}]$

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

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

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

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

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

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

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

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

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