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Semi transverse axis of hyperbola is 5. ...

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.

Text Solution

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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`
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