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Tangents are drawn to the hyperbola x^2/...

Tangents are drawn to the hyperbola `x^2/9-y^2/4=1` parallet to the sraight line `2x-y=1.` The points of contact of the tangents on the hyperbola are (A) `(9/(2sqrt2),1/sqrt2)` (B) `(-9/(2sqrt2),-1/sqrt2)` (C) `(3sqrt3,-2sqrt2)` (D) `(-3sqrt3,2sqrt2)`

A

`((9)/(2sqrt2),(1)/(sqrt2))`

B

`(-(9)/(2sqrt2),-(1)/(sqrt2))`

C

`(3sqrt3,-2sqrt2)`

D

`(3sqrt3,-2sqrt2)`

Text Solution

Verified by Experts

The correct Answer is:
A, B
Slope of tangent = 2
The tangent are
`y=2xpmsqrt(9xx4-4)" "("using y"=mx pmsqrt(a^(2)m^(2)-b^(2)))`
`"i.e., "2x-y= pm 4sqrt2`
`"or "(x)/(2sqrt2)-(y)/(4sqrt2)=1 and (x)/(2sqrt2)-(y)/(4sqrt2)=-1`
Comparing it with `(x x_(1))/(9)-(yy_(1))/(4)=1` (Eqn. of tangent to hypebola at point `(x_(1),y_(1))` on it we get the point of contact as `(9//2sqrt2,1//sqrt2) and (-9//2sqrt2, -1//sqrt2)`.
Alternatye Mathod:
The equation of tangent at `P(theta)` is
`((sec theta)/(3))x-((tan theta)/(2))y=1`
`therefore" Slope"=(2 sec theta)/(3 tan theta)=2`
`"or "sin theta=(1)/(3)`
`therefore" "sec theta= pm(3)/(2sqrt2)` and corresponding by tan `theta= pm (1)/(2sqrt2)`
Therefore, the points are `(9//2sqrt2,1//sqrt2)` and `(-9//2sqrt2, -1//sqrt2)`.
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