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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.

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.

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
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
askIITians Faculty

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