Let RS be the diameter of the circle x² + y² = 1, where S is the point (1, 0). Let P be a variable p

Q: Wait, why do we use parametric coordinates $P(\cos \theta, \sin \theta)$ instead of just $x$ and $y$?

Great question! While you could use Cartesian coordinates, the circle equation $x^2 + y^2 = 1$ is perfectly suited for trigonometry. Using $\theta$ as a parameter allows us to use powerful identities like the half-angle formulas, which turn complex algebraic expressions into simple, manageable ones. It is the most elegant path to the solution.

Q: How do I know the normal at P passes through the origin?

This is a fundamental property of circles! The normal at any point on a circle is always the line containing the radius at that point. Since the center of our circle is the origin $(0, 0)$, the normal must pass through the origin. This makes finding the equation of the normal as simple as finding the slope of the line connecting $(0, 0)$ and $P$.

Q: Why did we use the half-angle identity for $\tan \theta$?

We needed to eliminate the parameter $\theta$ to find the locus. By expressing $\tan \theta$ in terms of $\tan(\frac{\theta}{2})$, we created a direct algebraic link between the $x$ and $y$ coordinates of point E. Since $y = \tan(\frac{\theta}{2})$, this substitution allowed us to write the equation of the locus purely in terms of $x$ and $y$.

Question

Let $RS$ be the diameter of the circle $x^2 + y^2 = 1$, where $S$ is the point $(1, 0)$. Let $P$ be a variable point (other than $R$ and $S$) on the circle and tangents to the circle at $S$ and $P$ meet at the point $Q$. The normal to the circle at $P$ intersects a line drawn through $Q$ parallel to $RS$ at point $E$. Then the locus of $E$ passes through the point(s)

Options:

(A) $(\frac{1}{3}, \frac{1}{\sqrt{3}})$

(B) $(\frac{1}{4}, \frac{1}{2})$

(C) $(\frac{1}{3}, -\frac{1}{\sqrt{3}})$

(D) $(\frac{1}{4}, -\frac{1}{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a point $P$ moving along the unit circle defined by $x^2 + y^2 = 1$. We have a fixed diameter $RS$ lying on the x-axis, where $S = (1, 0)$ and $R = (-1, 0)$.

We define the position of $P$ using parametric coordinates as $P = (\cos 	heta, \sin 	heta)$. To avoid degenerate cases, we assume $P 
eq R, S$, which implies $	heta 
eq 0, \pi$.

### The Tangent Tango

The tangent line at $S$ is the vertical line $x = 1$. The tangent line at $P$ is given by the equation:
$$x \cos 	heta + y \sin 	heta = 1$$

To find the intersection point $Q$, we substitute $x = 1$ into the tangent equation for $P$:
$$(1) \cos 	heta + y \sin 	heta = 1$$

Solving for $y$, we obtain:
$$y_Q = \frac{1 - \cos 	heta}{\sin 	heta}$$

Using the half-angle identities $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$ and $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$, the expression simplifies to:
$$y_Q = 	an\left(\frac{	heta}{2}ight)$$

Thus, the coordinates of point $Q$ are $(1, 	an(\frac{	heta}{2}))$.

### The Normal and the Intersection

The normal to the circle at $P$ passes through the origin $(0, 0)$. Its slope is $\frac{\sin 	heta}{\cos 	heta} = 	an 	heta$, so the equation of the normal is:
$$y = x 	an 	heta$$

We also consider a line through $Q$ parallel to $RS$. Since $RS$ lies on the x-axis, this line is horizontal and defined by:
$$y = 	an\left(\frac{	heta}{2}ight)$$

Point $E$ is the intersection of these two lines. Setting the y-values equal, we have $x 	an 	heta = 	an(\frac{	heta}{2})$, which leads to:
$$x = \frac{	an(\frac{	heta}{2})}{	an 	heta}$$

### The Final Reveal

To find the locus, we eliminate $	heta$ using the double-angle identity $	an 	heta = \frac{2 	an(\frac{	heta}{2})}{1 - 	an^2(\frac{	heta}{2})}$. Substituting this into our expression for $x$:
$$x = \frac{	an(\frac{	heta}{2})}{\frac{2 	an(\frac{	heta}{2})}{1 - 	an^2(\frac{	heta}{2})}}$$

The $	an(\frac{	heta}{2})$ terms cancel out, simplifying the expression to:
$$x = \frac{1 - 	an^2(\frac{	heta}{2})}{2}$$

Since $y = 	an(\frac{	heta}{2})$, we substitute $y$ into the equation to get $x = \frac{1 - y^2}{2}$. Rearranging this yields:
$$y^2 = 1 - 2x$$

This is the equation of a **parabola**. By verifying the coordinates, we confirm the locus passes through points such as $(\frac{1}{3}, \frac{1}{\sqrt{3}})$ and $(\frac{1}{3}, -\frac{1}{\sqrt{3}})$.

Select Answer:

Visualized Solution

Visualizing the Circle and Diameter $R S$

Defining the Variable Point $P$

Equations of Tangents at $S$ and $P$

Finding the Intersection Point $Q$

Simplifying $Q$ using Half-Angle Formula

The Normal at $P$

Line through $Q$ Parallel to $R S$

Finding the Intersection Point $E$

Deriving the Locus of $E$ (Part 1)

Deriving the Locus of $E$ (Part 2)

Checking the Given Points

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Tango

The Normal and the Intersection

The Final Reveal

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Select Answer:

Visualized Solution

Visualizing the Circle and Diameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS

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

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

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q using Half-Angle Formula

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

Line through .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q Parallel to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Deriving the Locus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E (Part 1)

Deriving the Locus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E (Part 2)