Animated Solution for Mathematics - Conic Sections: 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 :

A Masterclass

Imagine you are standing before a parabola, $y^2=2x-3$. It is a simple, elegant curve, yet it holds secrets that only the keenest observers can unlock.

You are given an external point $R(0,1)$, and from this point, you draw two tangents that graze the parabola at points $P$ and $Q$. This is not just a problem; it is a dance of lines and curves. Our goal is to find the orthocenter of the triangle $PQR$.

The first step in our journey is to identify the chord of contact, the line segment $PQ$. Many students rush to find the equations of the tangents, but there is a more elegant path.

The equation of the chord of contact from an external point $(x_1,y_1)$ is given by $T=0$. For our parabola $y^2=2x-3$, this transformation is:

By substituting our point $R(0,1)$, we get $y(1)=(x+0)-3$, which simplifies beautifully to $y=x-3$. This linear equation is the backbone of our triangle.

Now that we have the chord $PQ$, we need to find the points $P$ and $Q$ where this line meets the parabola. We solve the system of equations $y=x-3$ and $y^2=2x-3$ simultaneously.

Substituting $y=x-3$ into the parabola equation, we get:

Expanding this, we arrive at $x^2-6x+9=2x-3$, which simplifies to the quadratic:

Factoring this, we find $(x-2)(x-6)=0$, giving us $x=2$ and $x=6$. These are the x-coordinates of our points $P$ and $Q$. Plugging these back into $y=x-3$, we find $P(2,-1)$ and $Q(6,3)$.

We now have the vertices of our triangle: $P(2,-1)$, $Q(6,3)$, and $R(0,1)$. A common mistake is to immediately start calculating the equations of the altitudes to find the orthocenter. But wait! Let us be smarter.

Let us calculate the slopes of the sides:

The product of these slopes is $1\cdot (-1)=-1$. This is the "Aha!" moment. The product of the slopes is $-1$, which means $PQ\perp PR$. The triangle $PQR$ is a right-angled triangle, with the right angle at $P$.

In any right-angled triangle, the orthocenter is the vertex containing the right angle. Since our right angle is at $P(2,-1)$, the orthocenter is simply $P(2,-1)$.

We have navigated the algebra and arrived at the geometric truth. This is the beauty of JEE problems—they reward not just calculation, but insight. You have successfully conquered this challenge.