Question

Prove that all the circles described on focal chords of the parabola as diameters, touch the directrix of the curve, and that all circles on focal radii as diameters, touch the tangent at the vertex.

Accepted Answer

Let the equation of the parabola be given as,

$y^{2} = 4 a x$

So, its focus $S$ , will be given as $(a, 0)$ .

Let $A S B$ , be any focal chord, where $A$ and $B$ are $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ , respectively.

So, the equation of $A B$ , is given as,

$y (t_{1} + t_{2}) = 2 x + 2 a t_{1} t_{2}$

As it passes through $(a, 0)$ , we get,

$2 a + 2 a t_{1} t_{2} = 0$ $\Rightarrow t_{1} t_{2} = - 1$

$∴$ $t_{2} = - \frac{1}{t_{1}}$

So, the coordinates of $B$ are $(\frac{a}{t_{1}^{2}}, \frac{- 2 a}{t_{1}})$ .

The equation of the circle drawn with $A B$ , as the diameter is given as,

$(x - a t_{1}^{2}) (x - \frac{a}{t_{1}^{2}}) + (y - 2 a t_{1}) (y + \frac{2 a}{t_{1}}) = 0 . . . (i)$

The equation of directrix is given as,

$x = - a . . . (i i)$

Solving equations $(i) & (i i)$ , simultaneously, we get,

$(- a - a t_{1}^{2}) (- a - \frac{a}{t_{1}^{2}}) + (y - 2 a t_{1}) (y + \frac{2 a}{t_{1}}) = 0$

$y^{2} - 2 a y (t_{1} - \frac{1}{t_{1}}) + a^{2} {(t_{1} - \frac{1}{t_{1}})}^{2} = 0$ $\Rightarrow {\{y - a (t_{1} - \frac{1}{t_{1}})\}}^{2} = 0$

As this is a perfect square, it will give us only one value of $y$ .

So, $(i i)$ cuts $(i)$ , at two coincident points.

Hence, the directrix of the parabola touches the circle drawn with the focal chord as the diameter.

Let the equation of circle drawn with the line $A S$ , as the diameter be given as,

$(x - a t_{1}^{2}) (x - a) + (y - 2 a t_{1}) (y - 0) = 0$

$\Rightarrow x^{2} + y^{2} - a x (1 + t_{1}^{2}) - 2 a t_{1} y + a^{2} {t_{1}}^{2} = 0 . . . (iii)$

The equation to the tangent at vertex is given as,

$x = 0 . . . (iv)$

Solving equations $(iii) & (iv)$ , we get,

$y^{2} - 2 a y t + a^{2} t_{1}^{2} = 0$ $\Rightarrow {(y - a t_{1})}^{2} = 0$

As this a perfect square, it will give us only one value of $y$ .

So, $(iii)$ cuts $(iv)$ , at only one point.

Hence, the tangent at the vertex of the parabola, touches the circle drawn on the focal radii as the diameter.

S L Loney Solutions for Chapter: Conic Sections. The Parabola, Exercise 4: EXAMPLES XXVIII

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: Conic Sections. The Parabola, Exercise 4: EXAMPLES XXVIII

Questions from S L Loney Solutions for Chapter: Conic Sections. The Parabola, Exercise 4: EXAMPLES XXVIII with Hints & Solutions

Learn and Prepare for any exam you want !