MHT CET · Maths · Ellipse
A common tangent to \(9 x^{2}-16 y^{2}=144\) and \(x^{2}+y^{2}=9\), is
- A \(y=\frac{3}{\sqrt{7}} x+\frac{15}{\sqrt{7}}\)
- 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
Answer & Solution
Correct Answer
(B) \(y=3 \sqrt{\frac{2}{7}} x+\frac{15}{\sqrt{7}}\)
Step-by-step Solution
Detailed explanation
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.
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.
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