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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced 2010 · Paper 1 · Mathematics · Hyperbola

The line 2 x+y=1 is tangent to the hyperbola...

The line 2x+y=12 x+y=1 is tangent to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. If this line passes through the point of intersection of the nearest directrix and the xx-axis, then the eccentricity of the hyperbola isThe line 2 x+y=1 is tangent to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 . If this line passes through the point ofThe line 2 x+y=1 is tangent to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 . If this line passes through the point of

Correct answer

2

Step-by-step solution

On substituting (ae,0)\left(\frac{a}{e}, 0\right) in y=-2 x+1, we get 0=−2ae+1⇒ae=12\begin{aligned}0 & =-\frac{2 a}{e}+1 \\ \Rightarrow \quad \frac{a}{e}& =\frac{1}{2} \end{aligned}Also, y=−2x+1y=-2 x+1 is tangent to hyperbola ∴1=4a2−b2⇒1a2=4−(e2−1)\begin{array}{ll} \therefore& 1=4 a²-b² \\ \Rightarrow & \frac{1}{a^2}=4-\left(e^2-1\right) \end{array}The line 2 x+y=1 is tangent to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 . If this line passes through the point of\\⇒4e2=5−e2⇒e4−5e2+4=0⇒(e2−4)(e2−1)=0⇒e=2,e=1\begin{array}{ll} \Rightarrow& \frac{4}{e^2}=5-e² \\ \Rightarrow & e⁴-5 e²+4=0 \\ \Rightarrow & \left(e^2-4\right) \left(e^2-1\right)=0 \\ \Rightarrow & e=2, e=1 \end{array}e=1e=1 gives the conic as parabola. But conic is given as hyperbola, hence $e=2
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