If the line y=mx+c is a common tangent to the hyperbola x^2100-y^264=1 and the circle x^2+…

Question

If the line $y=mx+c$ is a common tangent to the hyperbola $\dfrac{x^2}{100}-\dfrac{y^2}{64}=1$ and the circle $x^2+y^2=36$, then which one of the following is true?

Accepted Answer

$${{{x^2}} \over {100}} - {{{y^2}} \over {64}} = 1$$

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

$$ \Rightarrow $$ c = $$ \pm $$ $$\sqrt {100{m^2} - 64} $$

General tangent to hyperbola in slope form is

y = mx $$ \pm $$ $$\sqrt {100{m^2} - 64} $$

This tangent is also tangent to the circle x2
 + y2
 = 36, whose center (0, 0) and radius = 6.

Distance of the tangent from the center is

$$\left| {{{\sqrt {100{m^2} - 64} } \over {\sqrt {{m^2} + 1} }}} \right|$$ = 6

$$ \Rightarrow $$ 100m2
 --- 64 = 36m2
 + 36

$$ \Rightarrow $$ 64m2
 = 100

$$ \Rightarrow $$ m = $${{10} \over 8}$$

$$ \therefore $$ c2 = 100 $$ \times $$ $${{100} \over {64}}$$ - 64

$$ \Rightarrow $$ c2 = $${{{{100}^2} - {{64}^2}} \over {64}}$$

$$ \Rightarrow $$ c2 = $${{164 \times 36} \over {64}}$$

$$ \Rightarrow $$ 4c2 = 369

Solutions · 1