6.7 Q-1

Question Statement

Find the equations of the tangent and normal to the given conic at the indicated points.

i. $y^2 = 4a x$ at $(at^2, 2at)$
ii. $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(a \cos \theta, b \sin \theta)$
iii. $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(a \sec \theta, b \tan \theta)$

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Background and Explanation

This problem requires you to find the equations of tangents and normals to conic sections (parabolas, ellipses, and hyperbolas). To do this, we need to:

1. Differentiate the equation of the conic with respect to $x$ to find the slope of the tangent line at the given point.
2. Use the point-slope form of the equation of a line to find the equation of the tangent.
3. The slope of the normal line is the negative reciprocal of the tangent’s slope, and we use the point-slope form again to find the equation of the normal.

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Solution

i. $y^2 = 4a x$ at $(at^2, 2at)$

1. Differentiate the equation:

$$\frac{d}{dx}(y^2) = \frac{d}{dx}(4ax)$$

This gives:

$$2y \frac{dy}{dx} = 4a \quad \Rightarrow \quad \frac{dy}{dx} = \frac{2a}{y}$$

At point $(at^2, 2at)$, we substitute into the derivative:

$$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$$

So, the slope of the tangent is $\frac{1}{t}$, and the slope of the normal is $-t$.

2. Equation of the tangent:
   Using the point-slope form of the equation of a line:

$$y - 2at = \frac{1}{t}(x - at^2)$$

Simplifying:

$$y t - 2a^2 = x - at^2 \quad \Rightarrow \quad yt = x + at^2$$

So, the equation of the tangent is:

$$y t = x + at^2$$

3. Equation of the normal:
   The slope of the normal is $-t$, so using the point-slope form again:

$$y - 2at = -t(x - at^2)$$

Simplifying:

$$y - 2at = -t x + at^3 \quad \Rightarrow \quad y = -t x + at^3 + 2at$$

So, the equation of the normal is:

$$y = -t x + at^3 + 2at$$

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ii. $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(a \cos \theta, b \sin \theta)$

1. Differentiate the equation:

$$\frac{d}{dx} \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \right) = \frac{d}{dx}(1)$$

This gives:

$$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$$

Rearranging:

$$\frac{dy}{dx} = -\frac{b^2}{a^2} \frac{x}{y}$$

At $(a \cos \theta, b \sin \theta)$, we substitute:

$$\frac{dy}{dx} = -\frac{b^2}{a^2} \cdot \frac{a \cos \theta}{b \sin \theta} = \frac{-b \cos \theta}{a \sin \theta}$$

So, the slope of the tangent is $\frac{-b \cos \theta}{a \sin \theta}$, and the slope of the normal is $\frac{a \cos \theta}{b \sin \theta}$.

2. Equation of the tangent:
   Using the point-slope form:

$$y - b \sin \theta = \frac{-b \cos \theta}{a \sin \theta} (x - a \cos \theta)$$

Simplifying:

$$\frac{\sin \theta}{b}(y - b \sin \theta) = \frac{-\cos \theta}{a}(x - a \cos \theta)$$

This leads to:

$$\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$$

So, the equation of the tangent is:

$$\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$$

3. Equation of the normal:
   The equation of the normal is:

$$y - b \sin \theta = \frac{a \sin \theta}{b \cos \theta} (x - a \cos \theta)$$

Simplifying:

$$a \sin \theta x + b \cos \theta y = \sin \theta \cos \theta (a^2 - b^2)$$

So, the equation of the normal is:

$$a \sin \theta x + b \cos \theta y = \sin \theta \cos \theta (a^2 - b^2)$$

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iii. $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(a \sec \theta, b \tan \theta)$

1. Differentiate the equation:

$$\frac{d}{dx} \left( \frac{x^2}{a^2} - \frac{y^2}{b^2} \right) = \frac{d}{dx}(1)$$

This gives:

$$\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0$$

Rearranging:

$$\frac{dy}{dx} = \frac{b^2 x}{a^2 y}$$

At $(a \sec \theta, b \tan \theta)$, we substitute:

$$\frac{dy}{dx} = \frac{b^2 a \sec \theta}{a^2 b \tan \theta} = \frac{b \sec \theta}{a \tan \theta}$$

So, the slope of the tangent is $\frac{b \sec \theta}{a \tan \theta}$, and the slope of the normal is $-\frac{a \tan \theta}{b \sec \theta}$.

2. Equation of the tangent:
   Using the point-slope form:

$$y - b \tan \theta = \frac{b \sec \theta}{a \tan \theta} (x - a \sec \theta)$$

Simplifying:

$$\frac{x}{a} \sec \theta - \frac{y}{b} \tan \theta = 1$$

So, the equation of the tangent is:

$$\frac{x}{a} \sec \theta - \frac{y}{b} \tan \theta = 1$$

3. Equation of the normal:
   The equation of the normal is:

$$y - b \tan \theta = \frac{a \tan \theta}{b \sec \theta} (x - a \sec \theta)$$

Simplifying:

$$a \tan \theta x + b \sec \theta y = \sec \theta (a^2 - b^2)$$

So, the equation of the normal is:

$$a \tan \theta x + b \sec \theta y = \sec \theta (a^2 - b^2)$$

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Key Formulas or Methods Used

1. Tangent Slope Formula:

$$\frac{dy}{dx} = \text{Slope of tangent line}$$

2. Normal Slope:
   The slope of the normal is the negative reciprocal of the tangent slope:

$$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$$

3. Point-Slope Form:
   The equation of a line is given by:

$$y - y_1 = m(x - x_1)$$

where $m$ is the slope and $(x_1, y_1)$ is the point on the line.

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Summary of Steps

1. Differentiate the conic equation to find the slope of the tangent.
2. Use the point-slope form of the line to write the equation of the tangent.
3. Use the negative reciprocal of the tangent slope to find the slope of the normal.
4. Use the point-slope form again to write the equation of the normal.