MHT CET · Maths · Hyperbola
Focus of hyperbola is \((\pm 3,0)\) and equation of tangent is \(2 x+y-4=0\), find the equation of hyperbola.
- A \(4 x^{2}-5 y^{2}=20\)
- B \(5 x^{2}-4 y^{2}=20\)
- C \(4 x^{2}-5 y^{2}=1\)
- D \(5 x^{2}-4 y^{2}=1\)
Answer & Solution
Correct Answer
(A) \(4 x^{2}-5 y^{2}=20\)
Step-by-step Solution
Detailed explanation
Given, \((\pm a e, 0) =(\pm 3,0) \)
\( \Rightarrow a e =3 \)
\( \Rightarrow a^{2} e^{2} =9 \)
\( \Rightarrow b^{2}+a^{2} =9 \)
\( \because 2 x+y-4 =0 \)
\( \Rightarrow y=2 x+4\)
is the tangent to the hyperbola. \(\therefore\)
\((4)^{2}=a^{2}(-2)^{2}-b^{2}\)
\(\Rightarrow 4 a^{2}-b^{2}=16\)
On solving Eqs. (i) and (ii), we get \(a^{2}=5, b^{2}=4\)
\(\therefore\) Equation of hyperbola is \(\frac{x^{2}}{5}-\frac{y^{2}}{4}=1\)
\(\Rightarrow\) \(4 x^{2}-5 y^{2}=20\)
\( \Rightarrow a e =3 \)
\( \Rightarrow a^{2} e^{2} =9 \)
\( \Rightarrow b^{2}+a^{2} =9 \)
\( \because 2 x+y-4 =0 \)
\( \Rightarrow y=2 x+4\)
is the tangent to the hyperbola. \(\therefore\)
\((4)^{2}=a^{2}(-2)^{2}-b^{2}\)
\(\Rightarrow 4 a^{2}-b^{2}=16\)
On solving Eqs. (i) and (ii), we get \(a^{2}=5, b^{2}=4\)
\(\therefore\) Equation of hyperbola is \(\frac{x^{2}}{5}-\frac{y^{2}}{4}=1\)
\(\Rightarrow\) \(4 x^{2}-5 y^{2}=20\)
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