JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let a line \(L\) pass through the point of intersection of the lines \(b x+10 y-8=0\) and \(2 x-3 y=0\), \(b \in R -\left\{\frac{4}{3}\right\}\). If the line \(L\) also passes through the point \((1,1)\) and touches the circle \(17\left( x ^{2}+ y ^{2}\right)=16\), then the eccentricity of the ellipse \(\frac{x^{2}}{5}+\frac{y^{2}}{b^{2}}=1\) is.
- A \(\frac{2}{\sqrt{5}}\)
- B \(\sqrt{\frac{3}{5}}\)
- C \(\frac{1}{\sqrt{5}}\)
- D \(\sqrt{\frac{2}{5}}\)
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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