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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced 2011 · Paper 2 · Mathematics · Hyperbola

Let P(6,3) be a point on the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 ....

Let P(6,3)P(6,3) be a point on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. If the normal at the point PP intersects the XX-axis at (9,0)(9,0), then the eccentricity of the hyperbola is

Options

  1. A52\sqrt{\frac{5}{2}}
  2. B32\sqrt{\frac{3}{2}}
  3. C2\sqrt{2}
  4. D3\sqrt{3}

Correct answer

B. 32\sqrt{\frac{3}{2}}

Step-by-step solution

Equation of normal to hyperbola at (x1,y1)\left(x_1, y_1\right) is a2xx1+b2yy1=a2+b2∴ At (6,3),a2x6+b2y3=a2+b2Itpassesthrought\begin{gathered}\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b² \\ \therefore \text{ At }(6,3), \frac{a^2 x}{6}+\frac{b^2 y}{3}=a^2+b² \end{gathered} It passes throught(9,0).. Now, a2⋅96=a2+b2⇒3a22−a2=b2⇒a2b2=2∴e2=1+b2a2=1+12\Rightarrowe=32\begin{aligned}& \text{ Now, } \quad \frac{a^2 \cdot 9}{6}=a^2+b² \\ & \Rightarrow \quad \frac{3 a^2}{2}-a^2=b² \Rightarrow \frac{a^2}{b^2}=2 \\ & \therefore \quad e²=1+\frac{b^2}{a^2}=1+\frac{1}{2} \Rightarrowe=\sqrt{\frac{3}{2}} \end{aligned}$
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