AP EAMCET · Maths · Hyperbola
If a tangent to the hyperbola \(x^2-\frac{y^2}{3}=1\) is also a tangent to the parabola \(y^2=8 x\). then equation of such tangent with the positive slope is
- A \(y-x-\frac{1}{2}=0\)
- B \(y-2 x-1=0\)
- C \(2 y-4 x-1=0\)
- D \(y-x-1=0\)
Answer & Solution
Correct Answer
(B) \(y-2 x-1=0\)
Step-by-step Solution
Detailed explanation
Eq. of tangent to \(y^2=8 x\) is \(y=m x+\frac{2}{m}\) Eq. of tangent to \(x^2-\frac{y^2}{3}=1\) is \(y=m x \pm \sqrt{m^2-3}\) On comparing we get \(m= \pm 2\) \(\Rightarrow\) tangent with + ve slope is \(y-2 x-1=0\)
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