InterviewSolution
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InterviewSolution
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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) `(2/(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)` |
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Answer» Correct Answer - 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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