Q76 / 150 · MHT CET 2007 · Mathematics

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

  • A
    $y=\frac{3}{\sqrt{7}} x+\frac{15}{\sqrt{7}}$
  • Correct answer
    B
    $y=3 \sqrt{\frac{2}{7}} x+\frac{15}{\sqrt{7}}$
  • C
    $y=2 \sqrt{\frac{3}{7}} x+15 \sqrt{7}$
  • D
    None of the above

Solution

Let $y=m x+c$ be a common tangent to $9 x^{2}-16 y^{2}=144$ and $x^{2}+y^{2}=9 .$ Since, $y=m x+c \quad$ is a tangent to $9 x^{2}-16 y^{2}=144$ $\therefore c^{2}=a^{2} m^{2}-b^{2} \Rightarrow c^{2}=16 m^{2}-9 \quad \ldots$ (i) Now, $y=m x+c$ is a tangent to $x^{2}+y^{2}=9$ $\therefore \frac{c}{\sqrt{m^{2}+1}}=3 \Rightarrow c^{2}=9\left(1+m^{2}\right)$ From Eqs. (i) and (ii), we get $16 m^{2}-9=9+9 m^{2}$ $\Rightarrow \quad m=\pm 3 \sqrt{\frac{2}{7}}$ On putting the value of $m$ in $\mathrm{Eq}$ (ii), we get $c=\pm \frac{15}{\sqrt{7}}$ Hence, $y=3 \sqrt{\frac{2}{7}} x+\frac{15}{\sqrt{7}}$ is a common tangent.

Hence, the correct answer is option: 'B'.

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