Question
Through a point $P( - 2,0)$, tangents $PQ$ and $PR$ are drawn to the parabola $y^{2} = 8x$. Two circles each passing through the focus of the parabola and one touching parabola at $Q$ and other at $R$ are drawn. Which of the following point(s) with respect to the triangle $PQR$ lie(s) on the common chord of the two circles?
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Solution
Sol. [Key Concept: Each circle through the focus touching the parabola at a contact point passes through fixed centres of $\bigtriangleup \ PQR$; the common chord of the two circles passes through those centres.]
For y² = 8x, the tangent at (2t², 4t) is ty = x + 2t².
Since P(−2, 0) lies on the tangent, 0 = −2 + 2t² ⇒ t = ±1.
Hence, Q = (2, 4) and R = (2, −4).
A circle through focus (2, 0) and tangent to the parabola at Q is:
x² + y² − 8x − 4y + 12 = 0.
Similarly, the circle corresponding to R is:
x² + y² − 8x + 4y + 12 = 0.
Subtracting the two equations gives the common chord:y = 0.
Triangle PQR is an isosceles right triangle with P(−2, 0), Q(2, 4), R(2, −4).
Its centroid, orthocentre, incentre and circumcentre all lie on its axis of symmetry, y = 0.
Therefore, all four points lie on the common chord.
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