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MathHardAdv 2015 • Paper 2

Centroid Derivatives of Triangle Formed by Hyperbola Tangent

Let H:x2−y2=1H : x^2 - y^2 = 1 be a hyperbola and SS be a circle centered at N(x2,0)N(x_2, 0). Assume HH and SS are tangent to each other at point P(x1,y1)P(x_1, y_1) where x1>1x_1 > 1 and y1>0y_1 > 0. The common tangent to both curves at PP meets the xx-axis at point MM. If (l,m)(l, m) represents the centroid of triangle ΔPMN\Delta PMN, select the correct expression(s):

Options

A

dldx1=1−13x12\frac{dl}{dx_1} = 1 - \frac{1}{3x_1^2} for x1>1x_1 > 1

Correct
B

dmdx1=x13(x12−1)\frac{dm}{dx_1} = \frac{x_1}{3\left(\sqrt{x_1^2 - 1}\right)} for x1>1x_1 > 1

Correct
C

dldx1=1+13x12\frac{dl}{dx_1} = 1 + \frac{1}{3x_1^2} for x1>1x_1 > 1

D

dmdy1=13\frac{dm}{dy_1} = \frac{1}{3} for y1>0y_1 > 0

Correct

Topics & Concepts

Conic SectionsHyperbolaCirclesTangents and Normals

Step-by-Step Solution

To find the correct expressions for the derivatives of the centroid (l,m)(l, m) of triangle ΔPMN\Delta PMN, we analyze the geometry of the given hyperbola and circle step by step.

Step 1: Equation of the Normal and Center N(x2,0)N(x_2, 0)

The equation of the hyperbola is given by: H:x2−y2=1H : x^2 - y^2 = 1

Since the point P(x1,y1)P(x_1, y_1) lies on HH, we have: x12−y12=1  ⟹  y1=x12−1(since x1>1 and y1>0)x_1^2 - y_1^2 = 1 \implies y_1 = \sqrt{x_1^2 - 1} \quad (\text{since } x_1 > 1 \text{ and } y_1 > 0)

Differentiating x2−y2=1x^2 - y^2 = 1 implicitly with respect to xx: 2x−2ydydx=0  ⟹  dydx=xy2x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{y}

Thus, the slope of the tangent at P(x1,y1)P(x_1, y_1) is: mT=x1y1m_T = \frac{x_1}{y_1}

The slope of the normal to the hyperbola at P(x1,y1)P(x_1, y_1) is: mN=−y1x1m_N = -\frac{y_1}{x_1}

Since the circle SS centered at N(x2,0)N(x_2, 0) is tangent to HH at P(x1,y1)P(x_1, y_1), the line segment PNPN must lie along the normal to HH at PP. Therefore, the slope of PNPN is equal to mNm_N: 0−y1x2−x1=−y1x1\frac{0 - y_1}{x_2 - x_1} = -\frac{y_1}{x_1}

Since y1>0y_1 > 0, we can simplify this equation to: 1x2−x1=1x1  ⟹  x2−x1=x1  ⟹  x2=2x1\frac{1}{x_2 - x_1} = \frac{1}{x_1} \implies x_2 - x_1 = x_1 \implies x_2 = 2x_1

Thus, the coordinates of the center NN are: N=(2x1,0)N = (2x_1, 0)

Step 2: Coordinates of Point MM

The equation of the common tangent to HH at P(x1,y1)P(x_1, y_1) is given by: xx1−yy1=1x x_1 - y y_1 = 1

The point MM is the intersection of this tangent line with the xx-axis. Setting y=0y = 0: xx1=1  ⟹  x=1x1x x_1 = 1 \implies x = \frac{1}{x_1}

Thus, the coordinates of MM are: M=(1x1,0)M = \left(\frac{1}{x_1}, 0\right)

Step 3: Coordinates of Centroid (l,m)(l, m)

The vertices of triangle ΔPMN\Delta PMN are P(x1,y1)P(x_1, y_1), M(1x1,0)M\left(\frac{1}{x_1}, 0\right), and N(2x1,0)N(2x_1, 0).

The xx-coordinate of the centroid, ll, is: l=x1+1x1+2x13=3x1+1x13=x1+13x1l = \frac{x_1 + \frac{1}{x_1} + 2x_1}{3} = \frac{3x_1 + \frac{1}{x_1}}{3} = x_1 + \frac{1}{3x_1}

The yy-coordinate of the centroid, mm, is: m=y1+0+03=y13m = \frac{y_1 + 0 + 0}{3} = \frac{y_1}{3}

Step 4: Evaluating the Derivatives

  1. Derivative of ll with respect to x1x_1: dldx1=ddx1(x1+13x1)=1−13x12for x1>1\frac{dl}{dx_1} = \frac{d}{dx_1}\left(x_1 + \frac{1}{3x_1}\right) = 1 - \frac{1}{3x_1^2} \quad \text{for } x_1 > 1 This matches Option A.

  2. Derivative of mm with respect to x1x_1: Substituting y1=x12−1y_1 = \sqrt{x_1^2 - 1} into m=y13m = \frac{y_1}{3}: m=x12−13m = \frac{\sqrt{x_1^2 - 1}}{3} dmdx1=13⋅12x12−1⋅(2x1)=x13x12−1for x1>1\frac{dm}{dx_1} = \frac{1}{3} \cdot \frac{1}{2\sqrt{x_1^2 - 1}} \cdot (2x_1) = \frac{x_1}{3\sqrt{x_1^2 - 1}} \quad \text{for } x_1 > 1 This matches Option B.

  3. Derivative of mm with respect to y1y_1: dmdy1=ddy1(y13)=13for y1>0\frac{dm}{dy_1} = \frac{d}{dy_1}\left(\frac{y_1}{3}\right) = \frac{1}{3} \quad \text{for } y_1 > 0 This matches Option D.

Conclusion

The correct expressions are given by options A, B, and D.

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