Tangent and normal are drawn at P(16, 16) on the parabola y² = 16x, which intersect the axis of the

Q: Wait, why is the center of the circle the midpoint of $AB$?

Because $\angle APB = 90^\circ$ (since the tangent and normal are perpendicular). A circle passing through the vertices of a right-angled triangle has the hypotenuse as its diameter. Therefore, the center of the circumcircle must be the midpoint of the hypotenuse $AB$.

Q: Can I use the focus property to find $C$?

Yes! The center $C$ is $(4, 0)$, which is the focus of the parabola $y^2 = 16x$. This is a beautiful property of parabolas, but you can derive it using the midpoint formula just as easily.

Q: Why do we use the modulus in the $\tan \theta$ formula?

The angle between two lines is defined as the acute angle unless specified otherwise. The modulus ensures we get the positive value for the tangent of that acute angle, which is what we need.

Question

Tangent and normal are drawn at $P(16, 16)$ on the parabola $y^2 = 16x$, which intersect the axis of the parabola at $A$ and $B$, respectively. If $C$ is the centre of the circle through the points $P, A$ and $B$ and $\angle CPB = 	heta$, then a value of $	an 	heta$ is :

Options:

(A) 4/3

(B) 1/2

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the curve of the parabola $y^2 = 16x$. You pick a point $P(16, 16)$ and decide to draw two lines: a tangent, which kisses the curve, and a normal, which strikes it with absolute perpendicularity.

These two lines are the heartbeat of the parabola's geometry. As they race toward the axis of the parabola, they intersect the x-axis at points $A$ and $B$.

### Phase 1: The Tangent and Normal

First, let us identify our parameters. For the parabola $y^2 = 16x$, we have $4a = 16$, which implies $a = 4$.

The tangent at $P(16, 16)$ is given by the equation $yy_1 = 2a(x + x_1)$. Substituting our values:

$$16y = 8(x + 16) \Rightarrow 2y = x + 16$$

When this tangent hits the x-axis ($y=0$), we find $x = -16$. Thus, point $A$ is at $(-16, 0)$.

Now, consider the normal. The slope of our tangent is $m_T = \frac{1}{2}$. Since the normal is perpendicular, its slope must be $m_N = -2$.

Using the point-slope form, the equation of the normal is $y - 16 = -2(x - 16)$, which simplifies to:

$$y + 2x = 48$$

Setting $y=0$ to find where it cuts the x-axis, we get $2x = 48$, so $x = 24$. Point $B$ is at $(24, 0)$.

### Phase 2: The Circumcircle Insight

Because the tangent and normal are perpendicular, the angle $\angle APB$ is exactly $90^{\circ}$. This makes $	riangle PAB$ a right-angled triangle.

In geometry, a right-angled triangle inscribed in a circle has a special property: its hypotenuse is the diameter of that circle. Therefore, the circumcircle passing through $P, A,$ and $B$ has $AB$ as its diameter.

The center $C$ of this circle is simply the midpoint of $AB$. Calculating the midpoint of $A(-16, 0)$ and $B(24, 0)$:

$$C = \left( \frac{-16 + 24}{2}, 0 ight) = (4, 0)$$

### Phase 3: The Final Calculation

We need to find $	an 	heta$, where $	heta = \angle CPB$. We have the coordinates $C(4, 0)$, $P(16, 16)$, and $B(24, 0)$.

First, we find the slopes of the lines $CP$ and $PB$:

$$m_{CP} = \frac{16 - 0}{16 - 4} = \frac{16}{12} = \frac{4}{3}$$

$$m_{PB} = \frac{0 - 16}{24 - 16} = \frac{-16}{8} = -2$$

Now, we apply the formula for the tangent of the angle between two lines:

$$	an 	heta = \left| \frac{m_{CP} - m_{PB}}{1 + m_{CP} \cdot m_{PB}} ight|$$

Substituting our values:

$$	an 	heta = \left| \frac{\frac{4}{3} - (-2)}{1 + (\frac{4}{3})(-2)} ight| = \left| \frac{\frac{10}{3}}{1 - \frac{8}{3}} ight| = \left| \frac{\frac{10}{3}}{-\frac{5}{3}} ight| = |-2| = 2$$

The final result is **2**.

Select Answer:

Visualized Solution

The Parabola and Point $P$

Equation of Tangent

Finding Point $A$

Equation of Normal

Finding Point $B$

The Right Angle at $P$

Circumcircle and its Center $C$

Coordinates of Center $C$

Defining Angle $θ$

Slope of $C P$

Slope of $P B$

Angle Between Two Lines

Substituting Slopes

Simplifying the Expression

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

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

Equation of Tangent

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

Equation of Normal

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

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

Circumcircle and its Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Coordinates of Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

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

Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CP

Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PB