Let the normal at the point P on the parabola y² = 6x pass through the point (5, -8). If the tangent

Q: Why do we use the parametric form $P(at^2, 2at)$ instead of just $(x, y)$?

Using $(x, y)$ directly leads to complex algebraic expressions involving square roots or higher-degree polynomials. The parameter $t$ reduces the problem to a single variable, making the cubic equation manageable and the geometric interpretation much clearer.

Q: How do I know which root of the cubic equation to pick?

In JEE problems, there is typically one real root that satisfies the geometric constraints. In this case, $t=-2$ is the clear integer root that works perfectly. If you find multiple real roots, check if they correspond to different points on the parabola that satisfy the condition.

Q: Can I use the slope form of the normal $y = mx - 2am - am^3$?

Yes, you absolutely can! The slope form is equally valid. However, the parametric form is often more direct when dealing with a specific point $P$ on the parabola. Both methods will lead you to the same final result.

Question

Let the normal at the point $P$ on the parabola $y^2 = 6x$ pass through the point $(5, -8)$. If the tangent at $P$ to the parabola intersects its directrix at the point $Q$, then the ordinate of the point $Q$ is :

Options:

(A) -3

(B) $-\frac{9}{4}$

(C) $-\frac{5}{2}$

(D) -2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the parabola $y^2 = 6x$. It is a beautiful, symmetric curve, but it holds secrets that only reveal themselves when we apply the right tools.

Our journey begins by identifying the parameter $a$. By comparing $y^2 = 6x$ with the standard form $y^2 = 4ax$, we immediately find that $4a = 6$, which means:

$$a = \frac{3}{2}$$

This value of $a$ is the heartbeat of our parabola, defining its focus and its directrix.

### The Normal's Journey

We are looking for a point $P$ on this parabola such that the normal at $P$ passes through the external point $(5, -8)$. The normal is a line perpendicular to the tangent at $P$.

To handle this, we use the parametric coordinates $P(at^2, 2at)$. The equation of the normal at this point is a classic result:

$$y = -tx + 2at + at^3$$

This equation is our bridge. Since the normal passes through $(5, -8)$, this point must satisfy the equation. Substituting $x = 5$, $y = -8$, and $a = \frac{3}{2}$, we get:

$$-8 = -t(5) + 2\left(\frac{3}{2}\right)t + \left(\frac{3}{2}\right)t^3$$

Simplifying this, we get $-8 = -5t + 3t + \frac{3}{2}t^3$, which leads to $-8 = -2t + \frac{3}{2}t^3$. Multiplying by $2$ to clear the fraction, we arrive at the elegant cubic equation:

$$3t^3 - 4t + 16 = 0$$

By testing small integers, we find that $t = -2$ is a root, as $3(-8) - 4(-2) + 16 = -24 + 8 + 16 = 0$.

### The Tangent's Intersection

With $t = -2$, we can pinpoint $P$. The coordinates are $x = at^2 = \frac{3}{2}(-2)^2 = 6$ and $y = 2at = 2(\frac{3}{2})(-2) = -6$.

So, $P$ is at $(6, -6)$. Now, we need the tangent at $P$. The equation of the tangent to $y^2 = 4ax$ at $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.

Substituting $P(6, -6)$ and $2a = 3$, we get $y(-6) = 3(x + 6)$, which simplifies to $-2y = x + 6$, or:

$$x + 2y + 6 = 0$$

The problem asks for the intersection of this tangent with the directrix. The directrix of $y^2 = 4ax$ is $x = -a$, so for our parabola, it is $x = -\frac{3}{2}$.

Substituting $x = -\frac{3}{2}$ into our tangent equation, we get $-\frac{3}{2} + 2y + 6 = 0$. Solving for $y$, we find $2y = \frac{3}{2} - 6 = -\frac{9}{2}$, which gives:

$$y = -\frac{9}{4}$$

The ordinate of point $Q$ is therefore **$-\frac{9}{4}$**. We have navigated the geometry, solved the cubic, and found the intersection.

Animated Solution for Mathematics - Conic Sections: Let the normal at the point $P$ on the parabola $y^{2} = 6 x$ pass through the point $(5, - 8)$ . If the tangent at $P$ to the parabola intersects its directrix at the point $Q$ , then the ordinate of the point $Q$ is :

Select Answer:

Visualized Solution

Analyze the Parabola $y^{2} = 6 x$

Parametric Equation of Normal

Substitute the Point $(5, - 8)$

Simplify to a Cubic Equation

Solve for Parameter $t$

Locate Point $P$

Equation of Tangent at $P$

Directrix of the Parabola

Intersection Point $Q$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal's Journey

The Tangent's Intersection

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Select Answer:

Visualized Solution

Analyze the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=6x

Parametric Equation of Normal

Substitute the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (5,−8)

Simplify to a Cubic Equation

Solve for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Locate Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Equation of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P