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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

माना एक रेखा \(L\), रेखाओं \(bx +10 y -8=0\) तथा \(2 x -3 y =0, b \in R -\left\{\frac{4}{3}\right\}\) के प्रतिच्छेदन बिन्दु से होकर जाती है। यदि रेखा \(L\), बिन्दु \((1,1)\) से भी होकर जाती है तथा वृत्त \(17\left( x ^2+ y ^2\right)=16\) को स्पर्श करती है, तो दीर्घवृत्त \(\frac{x^2}{5}+\frac{y^2}{b^2}=1\) की उत्केन्द्रता है:

  1. A \(\frac{2}{\sqrt{5}}\)
  2. B \(\sqrt{\frac{3}{5}}\)
  3. C \(\frac{1}{\sqrt{5}}\)
  4. D \(\sqrt{\frac{2}{5}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{\frac{3}{5}}\)

Step-by-step Solution

Detailed explanation

Line is passing through intersection of \(b x+10 y-8=0\) and \(2 x-3 y=0\) is \((b x+10 y-8)+\lambda(2 x-3 y)=0\). As line is passing through \((1,1)\) so \(\lambda=b+2\) Now line \((3 b+4) x-(3 b-4) y-8=0\) is tangent to circle \(17\left(x^{2}+y^{2}\right)=16\) So…
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