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In the parabola y^2=4a x , then tangent ...

In the parabola `y^2=4a x ,` then tangent at `P` whose abscissa is equal to the latus rectum meets its axis at `T ,` and normal `P` cuts the curve again at `Qdot` Show that `P T: P Q=4: 5.`

Text Solution

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Let P be `(at^(2),2at)`. Since `at^(2)=4a,t=pm2`. Consider t=2
and P(4a,4a). The tangent at P 2y=x+4a which meets the x-axis at T(-4a,0).
If the coordinates of Q are `(at_(1)^(2),2at_(1))`, then
`t_(1)=-t-(2)/(t)=-3`
So, Q is (9a,-6a). Therefore,
`(PQ)^(2)=125a^(2)and(PT)^(2)=80a^(2)`
`orPT:PQ=4:5`
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Knowledge Check

  • If the tangent t the curv y=f(x )"at" P(x_(1),y_(1)) is paralle to the y-axis, then the of the normal to the curve at P is -

    A
    `y=x_(1)`
    B
    `y=y_(1)`
    C
    `x=x_(1)`
    D
    `x=y_(1)`

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