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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the equation of two diameters of a circle \(x ^{2}+ y ^{2}\) \(-2 x +2 fy +1=0\) be \(2 px - y =1\) and \(2 x + py =4 p\). Then the slope \(m \in(0, \infty)\) of the tangent to the hyperbola \(3 x^{2}-y^{2}=3\) passing through the centre of the circle is equal to \(......\)
- A \(6\)
- B \(2\)
- C \(4\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(2 p+f-1=0\) \(2-p f-4 p=0\) \(2=p(f+4)\) \(p=\frac{2}{f+4}\) \(2 p=1-f\) \(\frac{4}{f+4}=1-f\) \(f^{2}+3 f=0\) \(f=0 \text { or }-3\) Hyperbola \(3 x ^{2}- y ^{2}=3, x ^{2}-\frac{ y ^{2}}{3}=1\) \(y=m x \pm \sqrt{m^{2}-3}\) It passes \((1,0)\) \(o=m \pm \sqrt{m^{2}-3}\) \(m\)…
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