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Equation of Tangent at a Point (x,y) in Terms of f'(x)

A chord of a ...

Question

A chord of a parabola passes through a point on the axis(outside the parabola) whose distance from the vertex is half the latus rectum ; prove that the normals at its extremities meet on the curve.

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Solution

The equation of the parabola be $y^{2} = 4 a x$
Let the chord meets the parabola at $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$
Then equation of chord is
$(t_{1} + t_{2}) y = 2 x + 2 a t_{1} t_{2}$
It passes through $(- 2 a, 0)$
$(t_{1} + t_{2}) 0 = - 4 a + 2 a t_{1} t_{2} \Rightarrow t_{1} t_{2} = 2........ (i)$
Let the point of intersection of normals at $P$ and $Q$ be $(h, k)$
The point of intersection of normals at two given parametric point is
${2 a + a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2}), - a t_{1} t_{2} (t_{1} + t_{2})} {2 a + a ({(t_{1} + t_{2})}^{2} - t_{1} t_{2}), - a t_{1} t_{2} (t_{1} + t_{2})} {2 a + a ({(t_{1} + t_{2})}^{2} - t_{1} t_{2}), - a t_{1} t_{2} (t_{1} + t_{2})} = (h, k) h = 2 a + a ({(t_{1} + t_{2})}^{2} - t_{1} t_{2}) . . . . . . . . . . (i i) k = - a t_{1} t_{2} (t_{1} + t_{2})$
using $(i)$
$k = - 2 a (t_{1} + t_{2}) t_{1} + t_{2} = - \frac{k}{2 a}$
substituting in $(i i)$
$h = 2 a + a ({(t_{1} + t_{2})}^{2} - 2) h = a {(t_{1} + t_{2})}^{2} h = a \frac{k^{2}}{4 a^{2}} k^{2} = 4 a h$
generalising the equation
$y^{2} = 4 a x$
We get the same equation of parabola as the locus of point of intersection of tangents.
Hence proved.

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