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Grade: 12 

                        
find the eqution of locus of pt, the tangents from which to the parabola y square=32x have slopes m1 and m2 such that 1) m1-m2=8 2)m1/m2 - m2/m1=4
8 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points 
							Ans:
Hello Student,
Please find answer to your question

Let the Point P (h, k) from where two tangents are drawn to the parabola at points ‘t1’ and ‘t2’.
y^2 = 32x
y^{2} = 4(8)x
Points of intersection of two tangents:
h = 8t_{1}t_{2}
k = 8\frac{t_{1}+t_{2}}{2}
k = 4(t_{1}+t_{2})
Slope of the tangents:
\frac{1}{t_{1}},\frac{1}{t_{2}}
1)
\frac{1}{t_{1}} - \frac{1}{t_{2}} = 8
t_{2} - t_{1} = 8t_{1}t_{2}
(t_{2} - t_{1})^2 = (8t_{1}t_{2})^2
(t_{2} + t_{1})^2 - 4t_{1}t_{2} = (8t_{1}t_{2})^2
(\frac{k}{4})^2 - 4.\frac{h}{8} = (h)^2
\frac{k^2}{16} - \frac{8h}{16} = (h)^2
k^2 - 8h = 16h^2
y^2 - 8x = 16x^2
This is the locus.
2)
\frac{t_{2}}{t_{1}} - \frac{t_{1}}{t_{2}} = 4
t_{2}^2} - t_{1}^2 = 4t_{1}t_{2}
(t_{2} - t_{1})(t_{2} + t_{1}) = 4t_{1}t_{2}
(t_{2} - t_{1})^2(t_{2} + t_{1})^2 = 16t_{1}^2t_{2}^2
[(t_{2} + t_{1})^2-4t_{1}t_{2}](t_{2} + t_{1})^2 = 16t_{1}^2t_{2}^2
[(\frac{k}{4})^2-4.\frac{h}{8}](\frac{k}{4})^2 = 16(\frac{h}{8})^2
[\frac{k^2}{16}-\frac{8h}{16}]\frac{k^2}{16} = 16\frac{h}{64}
[k^2 - 8h]k^2 = 64h
[y^2 - 8x]y^2 = 64x
6 years ago
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