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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`
one year ago

```							 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).
```
one year ago
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