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The tangents from (1, 2sqrt2) to the hyp...

The tangents from `(1, 2sqrt2)` to the hyperbola `16x^2-25y^2 = 400` include between them an angle equal to:

Text Solution

Verified by Experts

The correct Answer is:
`y=-3xpmsqrt(209)`
We have hyperbola `(x^(2))/(25)-(y^(2))/(16)=1`.
The slope of given line `x-3y=4` is 1/3.
So, the slope of tangent is `-3.`
Therefore, equations of tangests are
`y=-3xpmsqrt(25xx9-16)`
`"or "y=-3x pm sqrt(209)`
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Knowledge Check

  • The coordinates of the vertices of the hyperbola 9x^(2)-16y^(2)=144 are-

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    `(7, -2sqrt(6))`
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    `(sqrt(6), 1)`

  • From the point (-1, -6) two tangents are drawn to the parabola y^(2)= 4x . Then the angle between the tangents is-

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    ` 30 ^(@)`
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