Consider the parabola y² = 4x. Let S be the focus of the parabola. A pair of tangents drawn to the p

Q: Wait, why do we use the slope form of the tangent $y = mx + a/m$ instead of the point-slope form?

Excellent question! While the point-slope form is valid, the slope form $y = mx + a/m$ is a 'power tool' for parabolas. It automatically embeds the condition of tangency. By substituting the point $P(-2, 1)$ into this equation, we instantly get a quadratic in $m$, which gives us both slopes simultaneously. It is the most elegant path to finding the two tangents.

Q: Is there a geometric shortcut to finding $SQ_1$ and $SQ_2$ without calculating the full line equations?

You have a sharp eye! There is a beautiful property: the angle between the focal radii $SP_1$ and $SP_2$ is related to the tangents. However, for a JEE Advanced problem, calculating the lines $SP_1$ and $SP_2$ is robust and prevents errors. Stick to the coordinate geometry method to ensure you don't miss any subtle signs or constants.

Q: Why is the distance $PQ_1$ equal to $PQ_2$ in this specific problem?

That is a consequence of the symmetry of the parabola and the specific location of point $P$. In this case, both perpendicular distances from $P$ to the focal lines turned out to be $3$. It is a delightful coincidence that simplifies our calculation of $SQ_1$ and $SQ_2$ significantly!

Q: I'm confused about the angle $\theta$ between $SP_1$ and $SP_2$. How do I calculate it correctly?

Think of the lines as vectors. The line $SP_1$ connects $(1, 0)$ to $(4, 4)$, giving it a direction vector $(3, 4)$. The line $SP_2$ connects $(1, 0)$ to $(1, -2)$, giving it a direction vector $(0, -2)$. Using the dot product formula $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|A||B|}$ is the most reliable way to find the angle between any two lines.

Q: Why do we use the Law of Cosines at the end?

Once we know $SQ_1 = 1$, $SQ_2 = 1$, and the angle $\theta$ between them, the triangle $\triangle SQ_1Q_2$ is fully defined. The Law of Cosines is the most direct way to find the side $Q_1Q_2$ when you have two sides and the included angle. It is the final, satisfying piece of the puzzle.

Question

Consider the parabola $y^2 = 4x$. Let $S$ be the focus of the parabola. A pair of tangents drawn to the parabola from the point $P = (-2, 1)$ meet the parabola at $P_1$ and $P_2$. Let $Q_1$ and $Q_2$ be points on the lines $SP_1$ and $SP_2$ respectively such that $PQ_1$ is perpendicular to $SP_1$ and $PQ_2$ is perpendicular to $SP_2$. Then, which of the following is/are TRUE ?

Options:

(A) $SQ_1 = 2$

(B) $Q_1Q_2 = \frac{3\sqrt{10}}{5}$

(C) $PQ_1 = 3$

(D) $SQ_2 = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a dark room, holding a flashlight at the point $P(-2, 1)$. Before you lies a parabolic mirror defined by the equation $y^2 = 4x$.

You shine your light, and two beams of light strike the parabola at points $P_1$ and $P_2$. These beams are the tangents. The focus $S(1, 0)$ of the parabola acts as a silent observer, and we are interested in the lines $SP_1$ and $SP_2$, the focal radii, and how they interact with your point $P$.

### The Quest for the Tangents

To begin our journey, we must define the paths of these light beams. For any parabola $y^2 = 4ax$, the equation of a tangent with slope $m$ is given by:

$$y = mx + \frac{a}{m}$$

With $a=1$, our equation simplifies to $y = mx + \frac{1}{m}$. Since these tangents must pass through our source $P(-2, 1)$, we substitute these coordinates into the equation:

$$1 = -2m + \frac{1}{m}$$

Multiplying by $m$ transforms this into the quadratic equation:

$$2m^2 + m - 1 = 0$$

Solving this, we find two distinct slopes: $m = \frac{1}{2}$ and $m = -1$. These are the slopes of our two light beams.

### Mapping the Points of Contact

With the slopes in hand, we locate the exact points where the light hits the mirror. Using the contact formula $P \equiv (\frac{a}{m^2}, \frac{2a}{m})$, we find:

$$P_1 = \left( \frac{1}{(1/2)^2}, \frac{2(1)}{1/2} ight) = (4, 4)$$

$$P_2 = \left( \frac{1}{(-1)^2}, \frac{2(1)}{-1} ight) = (1, -2)$$

Now, we connect these points to the focus $S(1, 0)$. The line $SP_1$ passes through $(1, 0)$ and $(4, 4)$, yielding the equation $4x - 3y - 4 = 0$. The line $SP_2$ is even simpler; since both $S$ and $P_2$ have an x-coordinate of $1$, the line is simply $x = 1$.

### The Perpendicular Projection

Now, we drop perpendiculars from $P(-2, 1)$ to these focal lines. Let $Q_1$ be the foot of the perpendicular on $SP_1$ and $Q_2$ on $SP_2$.

Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$, we calculate:

$$PQ_1 = \frac{|4(-2) - 3(1) - 4|}{\sqrt{4^2 + (-3)^2}} = \frac{|-8 - 3 - 4|}{5} = 3$$

Similarly, the distance from $P(-2, 1)$ to the vertical line $x=1$ is simply $|-2 - 1| = 3$.

### The Final Convergence

We are almost there. Consider the right-angled triangles $	riangle PSQ_1$ and $	riangle PSQ_2$. The hypotenuse is the distance $SP$, which is:

$$SP = \sqrt{(-2-1)^2 + (1-0)^2} = \sqrt{10}$$

By the Pythagorean theorem, $SQ_1 = \sqrt{SP^2 - PQ_1^2} = \sqrt{10 - 9} = 1$. Similarly, $SQ_2 = 1$.

To find the distance $Q_1Q_2$, we look at $	riangle SQ_1Q_2$. We have two sides of length $1$ and an included angle $	heta$. Using the direction vectors of the lines $SP_1$ and $SP_2$, we find $\cos 	heta = -\frac{4}{5}$.

Applying the Law of Cosines:

$$Q_1Q_2^2 = 1^2 + 1^2 - 2(1)(1)\left(-\frac{4}{5}ight) = 2 + \frac{8}{5} = \frac{18}{5}$$

Thus, the final distance is:

**$$Q_1Q_2 = \frac{3\sqrt{10}}{5}$$**

We have successfully navigated the geometry of the parabola, proving that the elegance of mathematics lies in the harmony of these calculated distances.

Select Answer:

Visualized Solution

Visual Anchor: The Setup

Logic Bridge: Equation of Tangent

Raw Setup: Passing through P

Atomic Compute: Finding Slopes

Atomic Compute: Points of Contact

Visual Anchor: Focal Radii

Atomic Compute: Equations of $S P_{1}$ and $S P_{2}$

Visual Anchor: Perpendiculars from P

Atomic Compute: Lengths of $P Q_{1}$ and $P Q_{2}$

Logic Bridge: Pythagoras Theorem

Atomic Compute: $S Q_{1}$ and $S Q_{2}$

Logic Bridge: Finding $Q_{1} Q_{2}$

Atomic Compute: Angle $θ$

Raw Setup: Cosine Rule

The Way Forward: Final Distance

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Quest for the Tangents

Mapping the Points of Contact

The Perpendicular Projection

The Final Convergence

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Select Answer:

Visualized Solution

Visual Anchor: The Setup

Logic Bridge: Equation of Tangent

Raw Setup: Passing through P

Atomic Compute: Finding Slopes

Atomic Compute: Points of Contact

Visual Anchor: Focal Radii

Atomic Compute: Equations of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SP1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SP2​

Visual Anchor: Perpendiculars from P

Atomic Compute: Lengths of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ2​

Logic Bridge: Pythagoras Theorem

Atomic Compute: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SQ1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SQ2​

Logic Bridge: Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q1​Q2​

Atomic Compute: Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Raw Setup: Cosine Rule

The Way Forward: Final Distance

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Quest for the Tangents

Mapping the Points of Contact

The Perpendicular Projection

The Final Convergence

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