Question

The line $y=mx+c$ becomes a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,then the value of c is

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

Answer 1

The correct option is B

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

For questions like these in which number of points of intersection are concerned we first solve the

equation .we get quadratic on any one variable .we then put determine as 0,+ve.or -ve according

to 1,2 and 0 number of solutions respectively .Solving equations of hyperbola and line

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

$x^2(a^2{m}^2-b^2)+\times \left(2a^{2}mc\right)+a^2(a^2+b^2)=0$.

Given that line touches the hyperbola $\therefore$ No.of solutions =1.

$\therefore$ Determinant = 0.

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

$e^2=a^2{m}^2-b^2$

$\therefore c=\pm \sqrt{a^2{m}^2-b^2}$ (option b)

Hence we get equations of line as,

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

Also for secant

$c^2>a^2{m}^2-b^2$

For line not $\frac{interesting}{touchinghyperbolaatall}$.

$c^2<a^2{m}^2-b^2$