The circle circumscribing the triangle formed by three tangents to a parabola passes through _____

Focus of the parabola

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Origin

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Vertex of the parabola

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Extremeties of latus rectum

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Solution

The correct option is A
Focus of the parabola

Let's take a standard parabola $y^{2} = 4 a x$

Let P, Q ad R be the points at which the tangents are drawn and

Let their coordinates be $(a t_{2} t_{3}, a (t_{2} + t_{3}))$

Similarly, the other pairs of tangents meet at the points

${a t_{3} t_{1}, a (t_{3} + t_{1})} a n d {a t_{1} t_{2}, a (t_{1} + t_{2})}$

Let the equation to the circle be

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ - - - - - - (1)

Since it passes through the above three points, we have

$a^{2} t_{2}^{2} t_{3}^{2} + a^{2} (t_{2} + t_{3})^{2} + 2 g a t_{2} t_{3} + 2 f a (t_{2} + t_{3}) + c = 0$ - - - - - - (2)

$a^{2} t_{3}^{2} t_{1}^{2} + a^{2} (t_{3} + t_{1})^{2} + 2 g a t_{3} t_{1} + 2 f a (t_{3} + t_{1}) + c = 0$ - - - - - - (3)

and $a^{2} t_{1}^{2} t_{2}^{2} + a^{2} (t_{1} + t_{2})^{2} + 2 g a t_{1} t_{2} + 2 f a (t_{1} + t_{2}) + c = 0$ - - - - - - (4)

Subtracting (3) from (2) and dividing by a $(t_{2} - t_{1})$ , we have

$a {t_{3}^{2} (t_{1} + t_{2}) + t_{1} + t_{2} + 2 t_{3}} + 2 g t_{3} + 2 f = 0$

Similarly, from (3) and (4), we have

$a (t_{1}^{2} (t_{2} + t_{3}) + t_{2} + t_{3} + 2 t_{1}) + 2 g t_{1} + 2 f = 0$

From these two equations we have

$2 g = - a (1 + t_{2} t_{3} + t_{3} t_{1} + t_{1} t_{2}) a n d 2 f = - a (t_{1} + t_{2} + t_{3} - t_{1} t_{2} t_{3})$

Substituting these values in (2), we obtain

$c = a^{2} (t_{2} t_{3} + t_{3} t_{1} + t_{1} t_{2})$

The equation to the circle is therefore

$x^{2} + y^{2} - a x (1 + t_{2} t_{3} + t_{3} t_{1} + t_{1} t_{2}) - a y (t_{1} + t_{2} + t_{3} - t_{1} t_{2} t_{3}) + a^{2} (t_{2} t_{3} + t_{3} t_{1} + t_{1} t_{2}) = 0$

Which clearly passes through the focus (a, 0)

Suggest Corrections

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