Question

The equations of the tangents to the hyperbola $x^2-4y^2=36$ which are perpendicular to the line $x-y+4=0$ are:

• A. $x+y+3\sqrt{3}=0$
• B. $x+y+3\sqrt{2}=0$
• C. $x+y-3\sqrt{3}=0$
• D. $x+y-3\sqrt{2}=0$

Answer 1

Answer: (A), (C)

$x+y-3\sqrt{3}=0$

$\mathrm{Let}m\mathrm{be}\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}.\mathrm{As}\mathrm{the}\mathrm{tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{line}x-y+4=0$
$So$, m$\times$1 = -1 $⟹$ m = -1
$\mathrm{Since},x^2-4y^2=36\mathrm{or}\frac{x^2}{36}-\frac{y^2}{9}=1\mathrm{So},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{tangents}\mathrm{are}:y=\left(-1\right)x\pm \sqrt{36\times (-1{)}^2-9}\Rightarrow y=-x\pm \sqrt{27}\Rightarrow y=-x\pm 3\sqrt{3}$