Through a point P are drawn tangents PQ and PR to a parabola and circles are drawn through the focus to touch the parabola in Q and R respectively; prove that the common chord of these circles passes through the centroid of the triangle PQR.

Open in App

Solution

Let the point of contact be $Q (a t_{1}^{2}, 2 a t_{1})$ and $R (a t_{2}^{2}, 2 a t_{2})$
Equation of tangent at $P$ is
$t_{1} y = x + a t_{1}^{2} x - t_{1} y + a t_{1}^{2} = 0$
Equation of circle through $Q$ and touching parabola is
${(x - a t_{1}^{2})}^{2} + {(y - 2 a t_{1})}^{2} + λ (x - t_{1} y + a t_{1}^{2}) = 0$
It passes through focus $(a, 0)$
$\Rightarrow λ = - a (1 + t_{1}^{2})$
So, the equation of circle is
${(x - a t_{1}^{2})}^{2} + {(y - 2 a t_{1})}^{2} + (- a (1 + t_{1}^{2})) (x - t_{1} y + a t_{1}^{2}) = 0 x^{2} + y^{2} - x (a + 3 a t_{1}^{2}) + y (a t_{1}^{3} - 3 a t_{1}) + 3 a^{2} t_{1}^{2} = 0...... (i)$
Similarly the equation of circle through $R$ is
$x^{2} + y^{2} - x (a + 3 a t_{2}^{2}) + y (a t_{2}^{3} - 3 a t_{2}) + 3 a^{2} t_{2}^{2} = 0...... (i i)$
Equation of common chord is obtained by subtracting two curves.
So equation of common chord is
$(i) - (i i) = 0$
$3 a x (t_{2}^{2} - t_{1}^{2}) + a y (t_{1}^{3} - 3 t_{1} - t_{2}^{3} + 3 t_{2}) + 3 a^{2} (t_{1}^{2} - t_{2}^{2}) = 0 3 a x {(t}_{2} + t_{1}) {(t}_{2} - t_{1}) + a y {{(t}_{1} - t_{2}) (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2}) - 3 {(t}_{1} - t_{2})} + 3 a^{2} {(t}_{1} - t_{2}) {(t}_{1} + t_{2}) = 0 - 3 a x {(t}_{2} + t_{1}) + a y (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} - 3) + 3 a^{2} {(t}_{1} + t_{2}) = 0...... (i i i)$
Point of intersection of tangents is $P (a t_{1} t_{2}, a (t_{1} + t_{2}))$
Centriod of $△ P Q R$ is
$(\frac{a t_{1}^{2} + a t_{2}^{2} + a t_{1} t_{2}}{3}, \frac{2 a t_{1} + 2 a t_{2} + a (t_{1} + t_{2})}{3}) (\frac{a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2})}{3}, a (t_{1} + t_{2}))$
Substituting in $(i i i)$ , we get
$- 3 a (\frac{a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2})}{3}) {(t}_{2} + t_{1}) + a (a) (t_{1} + t_{2}) (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} - 3) + 3 a^{2} {(t}_{1} + t_{2}) = 0 - a^{2} {(t}_{1}^{2} + t_{2}^{2} + t_{1} t_{2}) + a^{2} (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} - 3) + 3 a^{2} = 0 0 = 0$
Line satisfies the centroid of the triangle
Hence proved that centroid lies on the common chord of both the circles

Suggest Corrections

Similar questions

∠ R P Q = 30^{0}

∠ R Q S i n d e g r e e s

P

y^{2} = 4 x

P

T

Join BYJU'S Learning Program