InterviewSolution
Saved Bookmarks
| 1. |
Semi transverse axis of hyperbola is 5. Tangent at point P and normal to this tangent meet conjugate axis at A and B, respectively. The circle on AB as diameter passes through tow fixed points, the distance between which is 20. Find the eccentricity of hyperbola. |
|
Answer» Solution :CONSIDER hyperbola `(x^(2))/(a^(2))-(y^(2))/(B^(2))=1`. Tangent to it at any point `P(a sec theta, b tan theta)` is `(x)/(a) sec theta-(y)/(b) tan theta=1` It MEETS y-axis at `A(0,-b cot theta)`. NORMAL at point P is `ax cos theta+by cot theta=a^(2)e^(2)` It meets y-axis at `B(0,(a^(2)e^(2))/(b) tan theta)` Now, circle with diameter as AB is `x^(2)+(y+b cot theta)(y-(a^(2)e^(2))/(b)tantheta)=0` `rArr""x^(2)+y^(2)-(a^(2)e^(2))+(b cot theta-(a^(2)e^(2))/(b)tantheta)y=0` This circle passes through fixed point `(pm ae,0)`, DISTNACE between which is 2ae. `2ae=20"(given)"` `therefore""e=2` |
|