Q) The point on the hyperbola x224 - y218 = 1 which is nearest to the line 3x + 2y + 1 = 0 is
(A) (6,3)
(B) (-6,3)
(C) (6,-3)
(D) (-6,-3)
Acc. to me.. we take any point (asecθ , btanθ) on the hyperbola and find its perpendicular
distance from the line 3x + 2y + 1 = 0 and find its min. value.
But the problem is that i don't know how to find the min. value of the horrible expression.
(since we hv nt been taught max. and min.)
The solution given in my package is something like this...
Eqn. of the tangent at (√24 secθ, √18 tanθ) is
x secθ√24 - y tanθ√18 = 1
Since the point of contact is nearest to the line 3x + 2y + 1 = 0
Its slope = -3/2 => secθ/√24 x √18/tanθ = -3/2 => sinθ = -1/√3
Hence the point is (6,-3)
But since c2 < a2m2 - b2. so the point neither intersects nor touch the line.
So how can they make their slopes equal?
262since min. distance occurs only when the slope of the curve and the line are equal.
First of all..... I think u r a fiitjeean....right????? 😅😂..... I noted down as it is written in our package (page 7 PROBLEM 3)...pretty cool....as per ur Que.......tangent is drawn on the nearest point and it must have the same slope of that of the line....that's it .
I too have a Que.... from this method ,,after solving there arises two points (6,-3) && (-6,3).... so what should be the ans....ans if ans. Is (6,-3) then why not (-6,3).
-thanx
