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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.

Jayati Sindhu , 16 Years ago

Grade 12

1 Answers

Jitender Singh

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

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