1please solve further.
I am not getting any constant value.
I took any point as (4sin4θ,4cos4θ).
I got y intercept as 4cos2θ and x-intercept as 4tan2θ
Find the sum of the the intercepts on the axes of coordinates by any tangent to the curve √x +√y=2
Plz solve this problem............
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5 Answers
62take a pont h,k
find the value of dy/dx at h,k in terms of h and k
equation of the line will be
(y-k)=dy/dx(x-h)
The x intercept is given by substituting y=0 and similarly
now complete the proof. :)
1
1@ppphhh i think u ve done a calc mistake as x interceot cumes out to be 4sin2θ
11Answer is basically 4 units...(if i've not done any cal mistake)...
If point is (h,k), m=dy/dx=-(k/h)^(1/2)
X-intercept of tangent turns out as 2(h^{1/2})
Y-intercept = 2(k^1/2)
Adding 2(h^{1/2}+k^{1/2})=4units...
11slope of tangent at P(h,k) = -√(k/h)
equation of tang. at (h,k)is
y-k = -√(k/h)*(x-h)...(1)
the x-intercept=h+√kh (put y=0 in 1)
the y-intercept=k+√kh
adding x and y intercepts we get
= h+k+2√kh
=(√h+√k)2
=4 (from the equation of curve)