Consider the parabola with vertex (1/2, 3/4) and the directrix y = 1/2. Let P be the point where the

Q: Wait, why do we reject the negative root here?

Actually, we don't reject it! The quadratic equation gives two roots. One corresponds to $P$ (where we started) and the other to $Q$ (where the normal hits the parabola again). Both are valid, but we only need the one for $Q$ to find the distance.

Q: Why can't we just use the property of normals?

You absolutely can! But for JEE, mastering the algebraic derivation is safer and builds intuition. Let's stick to the fundamental method first to ensure you understand the geometry behind the formula.

Q: How do I know if my normal equation is correct?

Check the slope! If the tangent slope is $m_T$, the normal slope must be $-1/m_T$. If their product is $-1$, you're on the right track. Also, ensure the normal line passes through point $P$.

Question

Consider the parabola with vertex $(\frac{1}{2}, \frac{3}{4})$ and the directrix $y = \frac{1}{2}$. Let $P$ be the point where the parabola meets the line $x = -\frac{1}{2}$. If the normal to the parabola at $P$ intersects the parabola again at the point $Q$, then $(PQ)^2$ is equal to :

Options:

(A) $\frac{75}{8}$

(B) $\frac{125}{16}$

(C) $\frac{25}{2}$

(D) $\frac{15}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the beautiful world of coordinate geometry. Today, we are going to dissect a parabola, not just to solve a problem, but to understand its very DNA.

We start with a vertex $V = (\frac{1}{2}, \frac{3}{4})$ and a directrix $y = \frac{1}{2}$. Because the directrix is a horizontal line, we know immediately that our parabola opens vertically.

The focal length $a$ is the distance between the vertex and the directrix:

$$a = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$$

Using the standard form $(x - h)^2 = 4a(y - k)$, we substitute our values to obtain the governing equation:

$$(x - \frac{1}{2})^2 = 4\left(\frac{1}{4}\right)\left(y - \frac{3}{4}\right) \Rightarrow (x - \frac{1}{2})^2 = y - \frac{3}{4}$$

### The Point of Contact

Now, we are told that point $P$ lies on the line $x = -\frac{1}{2}$. To find its exact location, we substitute $x = -\frac{1}{2}$ into our parabola equation:

$$\left(-\frac{1}{2} - \frac{1}{2}\right)^2 = y - \frac{3}{4}$$

This simplifies to $(-1)^2 = y - \frac{3}{4}$, which implies $1 = y - \frac{3}{4}$, or $y = \frac{7}{4}$. Thus, our point $P$ is at $(-\frac{1}{2}, \frac{7}{4})$.

### The Normal's Path

To find the normal at $P$, we first need the slope of the tangent. We differentiate our parabola equation $(x - \frac{1}{2})^2 = y - \frac{3}{4}$ with respect to $x$:

$$2\left(x - \frac{1}{2}\right) = \frac{dy}{dx}$$

At $x = -\frac{1}{2}$, the slope of the tangent $m_T$ is:

$$m_T = 2\left(-\frac{1}{2} - \frac{1}{2}\right) = -2$$

The normal is perpendicular to the tangent, so its slope $m_N$ is:

$$m_N = -\frac{1}{m_T} = \frac{1}{2}$$

Using the point-slope form $y - y_1 = m(x - x_1)$, the equation of the normal is:

$$y - \frac{7}{4} = \frac{1}{2}\left(x + \frac{1}{2}\right) \Rightarrow y = \frac{x}{2} + 2$$

### The Intersection

The normal line intersects the parabola again at point $Q$. To find $Q$, we solve the system of equations $y = \frac{x}{2} + 2$ and $(x - \frac{1}{2})^2 = y - \frac{3}{4}$.

Substituting $y$ into the parabola equation:

$$(x - \frac{1}{2})^2 = \left(\frac{x}{2} + 2\right) - \frac{3}{4}$$

Expanding and rearranging terms leads to the quadratic equation:

$$x^2 - x + \frac{1}{4} = \frac{x}{2} + \frac{5}{4} \Rightarrow 2x^2 - 3x - 2 = 0$$

Factoring the quadratic gives $(2x + 1)(x - 2) = 0$. The roots are $x = -\frac{1}{2}$ (point $P$) and $x = 2$ (point $Q$). For $x = 2$, we find $y = \frac{2}{2} + 2 = 3$. Thus, $Q = (2, 3)$.

### Final Calculation

We now calculate the squared distance $(PQ)^2$ using the distance formula:

$$(PQ)^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$

$$(PQ)^2 = \left(2 - \left(-\frac{1}{2}\right)\right)^2 + \left(3 - \frac{7}{4}\right)^2 = \left(\frac{5}{2}\right)^2 + \left(\frac{5}{4}\right)^2$$

$$(PQ)^2 = \frac{25}{4} + \frac{25}{16} = \frac{100 + 25}{16} = \frac{125}{16}$$

The final result is **$\frac{125}{16}$**.

Select Answer:

Visualized Solution

Visualizing the Parabola Setup

Finding the Focal Length $a$

Standard Equation of the Parabola

Locating Point $P$

Coordinates of Point $P$

Finding the Slope of the Tangent

Slope of Tangent and Normal

Equation of the Normal Line

Intersection with the Parabola

Solving the Quadratic Equation

Finding the Coordinates of $Q$

Calculating $(P Q)^{2}$

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Point of Contact

The Normal's Path

The Intersection

Final Calculation

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