Tangents to y^2=4ax meets tangent at vertex at P and Q, if PQ=4, then show locus is y^2=8(x+2)

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Two tangents to parabola y^2=4ax meets tangent at vertex at P and Q, if PQ=4, then show that locus of point of intersection of the tangents is y^2=8(x+2) Support the channel:
  1. Mathsmerizing
  2. y^2=4ax
  3. length
  4. parametric
  5. tangent
  6. condition
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First of all, thank you so much for this to-the-point and elaborate explanation sir :)

Secondly, @3:52 the expression for |t1-t2| looks like the difference of roots of a quadratic equation ax^2+Beta.x+alpha=0. Does it have any link to that? Is that any method by which we can get the two points by forming that equation somehow?

kushagratiwari