JEE Advanced · Mathematics · 15. Hyperbola
Paragraph:
The circle \(x^2+y^2-8 x=0\) and hyperbola \(\frac{x^2}{9}-\frac{y^2}{4}=1\) intersect at the points \(A\) and \(B\).Question:
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
- A
\(2 x-\sqrt{5} y-20=0\)
- B
\(2 x-\sqrt{5} y+4=0\)
- C
\(3 x-4 y+8=0\)
- D
\(4 x-3 y+4=0\)
Answer & Solution
Correct Answer
(B)
\(2 x-\sqrt{5} y+4=0\)
Step-by-step Solution
Detailed explanation
Equation of tangent to hyperbola having slope \(m\) is
\[
y=m x+\sqrt{9 m^2-4}
\]
Equation of tangent to circle is
\[
y=m(x-4)+\sqrt{16 m^2+16}
\]
Eqs. (i) and (ii) will be identical for \(m=\frac{2}{\sqrt{5}}\) satisfy.
\(\therefore\) Equation of common tangent is \(2 x-\sqrt{5} y+4=0\)
\[
y=m x+\sqrt{9 m^2-4}
\]
Equation of tangent to circle is
\[
y=m(x-4)+\sqrt{16 m^2+16}
\]
Eqs. (i) and (ii) will be identical for \(m=\frac{2}{\sqrt{5}}\) satisfy.
\(\therefore\) Equation of common tangent is \(2 x-\sqrt{5} y+4=0\)
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