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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 10:14:45 IST
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if m1 n m2 are slopes of tangents frm a point (1,4) on 16x2 - 25y2 = 400, then the point frm which the tangents drawn on hyperbola have slope | m1 | n | m2 | n positive intercept on Y-axis is (a) (-7, -4) (b) (7, 4) (c) (-4, -7) (d) (4,7) firstly pls tell how do u find a point frm where slopes of both tangents to the hyperbola are positive?? if ur considerin the 2 diff hyperbolas represented by the same eqn then its possible bt if its just one continuous hyperbola to right or left of the conjugate axis, then how do gt both tangents wit positive slopes??
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29 Nov 2007 12:27:45 IST
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hyperbola has to parts, on each side of the conjugate axis
visualize u draw two tangents to it frm the origin, touching the hyp at a point in the 1st quadrant and the other in the 3rd quadrant, then the two tangents will have positive slopes
this was 4 ur first query
lemme try the question
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29 Nov 2007 12:33:18 IST
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is the answer a
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- Edward Everett Hale
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eqn of tangent is
y=mx+/- rt(25m^2-16)
putting (1, 4) here, we get two values of m as 1 and -4/3
so tangents with positive slopes will have m as 1 and 4/3
again put'em in y=mx+rt(25m^2-16)
dis time take only +sign 4 intercept
u'll get eqns of tangent as y=x+3 n y=4/3x+16/3
dere point of intersctin is (-7,-4)
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
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I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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29 Nov 2007 13:45:42 IST
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bulls eye..gr8 answer..
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