Question

What is the condition for a line $y=mx+c$ to be tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

• A. $c=\pm \sqrt{b^2-a^2{m}^2}$
• B. $c=\pm \sqrt{a^2{m}^2-b^2}$
• C. $c=\pm \sqrt{a^2-b^2{m}^2}$
• D. $c=\pm \sqrt{b^2+a^2{m}^2}$

Answer 1

The correct option is A

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

A line $y=mx+c$ becomes a tangent to a curve when on solving the equations of the line and curve we get one solution.

Here solving these equation gives,

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

$b^2{x}^2-a^2(mx+c{)}^2=a^2{b}^2$

$x^2.(b^2-a^2{m}^2)-2a^{2}mcx-a^2{c}^2-a^2{b}^2=0$

We get one solution if $\Delta =0.$

i.e.,$4a^4{m}^2{c}^2+4(b^2-a^2{m}^2)a^2(c^2-b^2)=0$

$a^2{m}^2{c}^2+(b^2{c}^2-64-a^2{m}^2{c}^2+a^2{b}^2{m}^2)=0$

$c^2-b^2+a^2{m}^2=0$

$\therefore c^2=b^2-a^2{m}^2.$

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