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find the eqution of locus of pt, the tangents from which to the parabola y square=32x have slopes m1 and m2 such that1) m1-m2=82)m1/m2 - m2/m1=4

pooja rathi , 12 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

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)
$\frac{t_{2}}{t_{1}} - \frac{t_{1}}{t_{2}} = 4$
$t_{2}^2} - t_{1}^2 = 4t_{1}t_{2}$
$(t_{2} - t_{1})(t_{2} + t_{1}) = 4t_{1}t_{2}$
$(t_{2} - t_{1})^2(t_{2} + t_{1})^2 = 16t_{1}^2t_{2}^2$
$[(t_{2} + t_{1})^2-4t_{1}t_{2}](t_{2} + t_{1})^2 = 16t_{1}^2t_{2}^2$
$[(\frac{k}{4})^2-4.\frac{h}{8}](\frac{k}{4})^2 = 16(\frac{h}{8})^2$
$[\frac{k^2}{16}-\frac{8h}{16}]\frac{k^2}{16} = 16\frac{h}{64}$
$[k^2 - 8h]k^2 = 64h$
$[y^2 - 8x]y^2 = 64x$