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The parabolas y2=4ax and x2=4by intersect orthogonally at point P(x1,y1) where x1,y1=0 is not possible, then

(A)b=a2
(B)b=a3
(C)b3=a2
(D)None of these.
7 years ago

Jitender Singh
IIT Delhi
158 Points

Ans:
If two parabolas intersect orthogonally, then the tangents at points of intersection must be perpendicular. Lets find points of intersection:
$y^{2}=4ax$...(1)
$x^{2}=4by$...(2)
Put y from (2) in (1)
$(\frac{x^{2}}{4b})^{2} = 4ax$
$x_{0} = 4a^{1/3}b^{2/3}, 0$
$y_{0} = 4a^{2/3}b^{1/3}, 0$
Let the slope of tangent to parabola (1) & (2) be m1& m2respectively.
$m_{1}.m_{2} = -1$
$m_{1} = \frac{2a}{y_{0}}$
$m_{2} = \frac{x_{0}}{2b}$
$\frac{2a}{y_{0}}. \frac{x_{0}}{2b}=-1$
$\frac{a.a^{1/3}.b^{2/3}}{b.b^{1/3}.a^{2/3}} = -1$
$\frac{a^{2/3}}{b^{2/3}} = -1$
Thanks & Regards
Jitender Singh
IIT Delhi
3 years ago
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