JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Given; A circle \(2{x^2} + 2{y^2} = 5\) and parabola \({y^2} = 4\sqrt 5 x\) Statement \(-1\):An equation of a common tangent to these curve is \(y = x + \sqrt 5 \) Statement \(-2\): If the line, \(y = mx + \frac{{\sqrt 5 }}{m}\left( {m \ne 0} \right)\) is their common tangent , then \(m\) satisfies \({m^4} - 3{m^2} + 2 = 0\).
- A Statement \(-1\) is false, Statement \(-2\) is true
- B Statement \(-1\) is true, Statement \(-2\) is false
- C Statement \(-1\) is true, Statement \(-2\) is true; Statement \(-2\) is acorrect explanation for Statement \(-1\)
- D Statement \(-1\) is true, Statement \(-2\) is true; Statement \(-2\) is not acorrect explanation for Statement \(-1\)
Answer & Solution
Correct Answer
(D) Statement \(-1\) is true, Statement \(-2\) is true; Statement \(-2\) is not acorrect explanation for Statement \(-1\)
Step-by-step Solution
Detailed explanation
Let common tangent \( y=m x +\frac{\sqrt{5}}{m} \) \(\frac{{\frac{{\sqrt 5 }}{m}}}{{\sqrt {1 + {m^2}} }} = \sqrt {\frac{5}{2}} \) \( m \sqrt{1+m^{2}}=\sqrt{2} \) \( m^{2}\left(1+m^{2}\right) =2 \) \( m^{4}+m^{2}-2 =2 \) \(\left(m^{2}+2\right)\left(m^{2}-1\right) =0 \)…
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