Question

# Show that the locus of the intersection of two perpendicular normals to an ellipse is the curve (a2+b2)(x2+y2)(a2y2+b2x2)2=(a2−b2)2(a2y2−b2x2)2.

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Solution

## As the normals are perpendicular, therefore the corresponding tangents are also perpendicular. So they must intersect on the director circle which is given by x2+y2=a2+b2=A2 (say)Any point P on the director circle may be taken as, (Acosθ,Asinθ)Let the equation to the ellipse be x2a2+y2b2=1So the polar of P w.r.t. the ellipse will be Axcosθa2+Aysinθb2=1The intersection of the perpendicular normals may be given by the equationSo putting l=Acosθa2 and m=Asinθb2 we get, x=(a2−b2)Acosθa21−(a2+b2)b2sin2θa2+b2a2−cos2θ+⎧⎪⎨⎪⎩(a2+b2)sin2θb2⎫⎪⎬⎪⎭∴x=a2−b2a2+b2b2cos2θ−a2sin2θb2cos2θ+a2sin2θAcosθ....1and y=−(a2−b2)Asinθb21−(a2+b2a2)cos2θ(a2+b2)a2cos2θ+(a2+b2)b2sin2θ=a2−b2a2+b2b2cos2θ−a2sin2θb2cos2θ+a2sin2θAsinθ.....2xy=cosθsinθsinθy=cosθx=1√x2+y2.....3Eliminating from θ from 1 and 3, we getx=a2−b2a2+b2Ax√x2+y2b2x2−a2y2b2x2+a2y2or (a2+b2)2(x2+y2)(b2x2+a2y2)2=(a2−b2)2(b2x2−a2y2)2

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