A common tangent to 9x^{2}-16y^{2}=144 and x^ | vedclass

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Question

(B) Let $y=mx+c$ be a common tangent to the hyperbola $9x^{2}-16y^{2}=144$ and the circle $x^{2}+y^{2}=9$. Rewriting the hyperbola as $\frac{x^{2}}{16

Vedclass · Accepted Answer

$y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$

Vedclass · Answer

(B) Let $y=mx+c$ be a common tangent to the hyperbola $9x^{2}-16y^{2}=144$ and the circle $x^{2}+y^{2}=9$. Rewriting the hyperbola as $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$,we have $a^{2}=16$ and $b^{2}=9$. The condition for $y=mx+c$ to be a tangent to the hyperbola is $c^{2}=a^{2}m^{2}-b^{2}$,so $c^{2}=16m^{2}-9$ $(i)$. The condition for $y=mx+c$ to be a tangent to the circle $x^{2}+y^{2}=3^{2}$ is $c^{2}=r^{2}(1+m^{2})$,so $c^{2}=9(1+m^{2})$ (ii). Equating $(i)$ and (ii): $16m^{2}-9=9+9m^{2}$. $7m^{2}=18$ $\Rightarrow m^{2}=\frac{18}{7}$ $\Rightarrow m=\pm 3\sqrt{\frac{2}{7}}$. Substituting $m^{2}=\frac{18}{7}$ into (ii): $c^{2}=9(1+\frac{18}{7})=9(\frac{25}{7})=\frac{225}{7}$. Thus,$c=\pm \frac{15}{\sqrt{7}}$. The common tangent is $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$.

$A$ common tangent to $9x^{2}-16y^{2}=144$ and $x^{2}+y^{2}=9$ is

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