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two equal parabolas have the same focus & their axis at right angles, a normal to one is perpendicular to the normal to the other.Prove that the locus of the intersection of these normals is another parabola
The equation of the normal to the parabola y2 = 4ax is y = mx – 2am – am3.It passes through the point (h, k) if k = mh – 2am – am3 => am3 + m(2a – h) + k = 0. … (1)Let the roots of the above equation be m1, m2and m3. Let the perpendicular normals correspond to the values of m1 and m2 so that m1m2 = –1.From equation (1), m1 m2 m3 = -k/a. Since m1 m2 = –1, m3 = k/a.Since m3 is a root of (1), we have a (k/a)3+k/a (2a – h) + k = 0. ⇒ k2 + a(2a – h) + a2 = 0⇒ k2 = a(h – 3a).Hence the locus of (h, k) is y2 = a(x – 3a).
The equation of the normal to the parabola y2 = 4ax is y = mx – 2am – am3.
It passes through the point (h, k) if
k = mh – 2am – am3 => am3 + m(2a – h) + k = 0. … (1)
Let the roots of the above equation be m1, m2and m3. Let the perpendicular normals correspond to the values of m1 and m2 so that m1m2 = –1.
From equation (1), m1 m2 m3 = -k/a. Since m1 m2 = –1, m3 = k/a.
Since m3 is a root of (1), we have a (k/a)3+k/a (2a – h) + k = 0. ⇒ k2 + a(2a – h) + a2 = 0
⇒ k2 = a(h – 3a).
Hence the locus of (h, k) is y2 = a(x – 3a).
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