If the tangents drawn at the points P and Q o | vedclass

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Question

(B) The equation of the parabola is $y^{2} = 2x - 3$. The equation of the chord of contact $PQ$ for the point $R(0, 1)$ is given by $T = 0$. $y(1) = 1

Vedclass · Accepted Answer

$(2, -1)$

Vedclass · Answer

(B) The equation of the parabola is $y^{2} = 2x - 3$. The equation of the chord of contact $PQ$ for the point $R(0, 1)$ is given by $T = 0$. $y(1) = 1(x + 0) - 3 \implies y = x - 3$. Substituting $x = y + 3$ into the parabola equation: $y^{2} = 2(y + 3) - 3 = 2y + 6 - 3 = 2y + 3$. $y^{2} - 2y - 3 = 0 \implies (y - 3)(y + 1) = 0$. Thus,$y = 3$ or $y = -1$. For $y = 3$,$x = 6$. For $y = -1$,$x = 2$. So,the points are $P(2, -1)$ and $Q(6, 3)$ (or vice versa). The slope of $PQ$ is $m_{PQ} = \frac{3 - (-1)}{6 - 2} = \frac{4}{4} = 1$. The slope of $PR$ is $m_{PR} = \frac{1 - (-1)}{0 - 2} = \frac{2}{-2} = -1$. Since $m_{PQ} 	imes m_{PR} = 1 	imes (-1) = -1$,the triangle $PQR$ is a right-angled triangle at $P$. In a right-angled triangle,the orthocentre is the vertex where the right angle is formed. Therefore,the orthocentre is $P(2, -1)$.

If the tangents drawn at the points $P$ and $Q$ on the parabola $y^{2} = 2x - 3$ intersect at the point $R(0, 1)$,then the orthocentre of the triangle $PQR$ is.

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