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  • A common tangent of <img alt="\mathrm{9 x^{2}-16 y^{2}=144\: and \: x^{2}+y^{2}=9}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B9%20x%5E%7B2%7D-16%20y%5E%7B2%7D%3D144%5C%3A%20a

A common tangent of \mathrm{9 x^{2}-16 y^{2}=144\: and \: x^{2}+y^{2}=9} is

Option: 1

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


Option: 2

\mathrm{y} \sqrt{7}=3 \mathrm{x} \sqrt{5}-15


Option: 3

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


Option: 4

None of these


Answers (1)

best_answer

Any tangent to \mathrm{\frac{x^{2}}{16}-\frac{y^{2}}{9}=1} is \mathrm{y=m x+c} where, \mathrm{c^{2}=16 m^{2}-9}.

If it touch the circle \mathrm{x^{2}+y^{2}=9}, then
\mathrm{ \frac{|c|}{\sqrt{1+m^{2}}}=3}
\mathrm{ \Rightarrow \mathrm{c}^{2}=9\left(1+\mathrm{m}^{2}\right)=16 \mathrm{~m}^{2}-9 \Rightarrow \mathrm{m}^{2}=\frac{18}{7}}
\mathrm{ \Rightarrow c^{2}=16 \cdot \frac{18}{7}-9=\frac{225}{7}}

\mathrm{ \Rightarrow} Hence, common tangents are
\mathrm{ y \pm 3 x \sqrt{\frac{2}{7}} \pm \frac{15}{\sqrt{7}}}
\mathrm{ \text { i.e. } y \sqrt{7}= \pm 3 x \sqrt{2} \pm 15}

Posted by

Irshad Anwar

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