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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 & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
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