1take two points θ1 and θ2... and proceed..
(bad suggestion though.. the process will be lengthy.. will tell if cud get a shorter one..)
equation of locus of point of intersection of 2 tangents of d hyperbola
x2/a2-y2/b2 =0
such dat d sums of their slopes is 5
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UP 0 DOWN 0 0 3
3 Answers
11No man thats a bad idea.
I know a quicker one.
We know the condition for tangency,
c=±√a2m2+b2
Substituting in y=mx+c,
lets rearrange the equation to get a quadratic in m.
m2(x2-a2)-2xym+y2+b2=0
From the theory of quadratic equations we have,
Let roots be m1 and m2.
Now m1+m2=2xy/(x2-a2)
Now put m1+m2=5,
And we get,
5x2-5a2-y2-b2=0
Voila !!
You locus is ready.
:)
