6.2 Q-1

Question Statement

The problem asks to write down the equations of the tangent and normal lines to two different circles at specific points:

1. For the circle $x^2 + y^2 = 25$, find the equations of the tangent and normal at:

   • $(4, 3)$
   • $(5 \cos \theta, 5 \sin \theta)$

2. For the ellipse $3x^2 + 2y^2 + 5x - 13y + 2 = 0$, find the equations of the tangent and normal at the point $\left( 1, \frac{10}{3} \right)$.

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Background and Explanation

Before solving this, let’s review some key concepts:

• Equation of a Tangent to a Circle: The equation of the tangent at a point on the circle can be found using the derivative of the circle’s equation, or by using the point-slope form of the line.

• Slope of Tangent Line: For a circle given by $x^2 + y^2 = r^2$, the slope of the tangent line at a point $(x_1, y_1)$ is given by:

$$m = -\frac{x_1}{y_1}$$

• Equation of Normal Line: The normal to the circle is perpendicular to the tangent. The slope of the normal line is the negative reciprocal of the tangent’s slope.

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Solution

i. Circle $x^2 + y^2 = 25$

Step 1: Find the slope of the tangent at $(4, 3)$
The equation of the circle is $x^2 + y^2 = 25$. To find the slope of the tangent at $(x_1, y_1)$, we differentiate implicitly:

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

Solving for $\frac{dy}{dx}$ (the slope of the tangent):

$$\frac{dy}{dx} = -\frac{x}{y}$$

At the point $(4, 3)$:

$$\frac{dy}{dx} = -\frac{4}{3}$$

Thus, the slope of the tangent at $(4, 3)$ is $-\frac{4}{3}$.

Step 2: Find the equation of the tangent at $(4, 3)$
Using the point-slope form of the line $y - y_1 = m(x - x_1)$, where $m$ is the slope:

$$y - 3 = -\frac{4}{3}(x - 4)$$

Simplifying:

$$3y - 9 = -4x + 16 \quad \Rightarrow \quad 4x + 3y - 25 = 0$$

Thus, the equation of the tangent at $(4, 3)$ is $4x + 3y - 25 = 0$.

Step 3: Find the equation of the normal at $(4, 3)$
The slope of the normal is the negative reciprocal of the slope of the tangent:

$$m_{\text{normal}} = \frac{3}{4}$$

Using the point-slope form again:

$$y - 3 = \frac{3}{4}(x - 4)$$

Simplifying:

$$4y - 12 = 3x - 12 \quad \Rightarrow \quad 3x - 4y = 0$$

Thus, the equation of the normal at $(4, 3)$ is $3x - 4y = 0$.

Step 4: Equation of the tangent at $(5 \cos \theta, 5 \sin \theta)$
Using the slope formula for the tangent at any point on the circle $x^2 + y^2 = 25$:

$$\frac{dy}{dx} = -\frac{x}{y} = -\frac{5 \cos \theta}{5 \sin \theta} = -\frac{\cos \theta}{\sin \theta}$$

So, the slope of the tangent at $(5 \cos \theta, 5 \sin \theta)$ is $-\frac{\cos \theta}{\sin \theta}$.

Using the point-slope form of the tangent:

$$y - 5 \sin \theta = \frac{\cos \theta}{\sin \theta}(x - 5 \cos \theta)$$

Simplifying:

$$y \sin \theta - 5 \sin^2 \theta = -x \cos \theta + 5 \cos^2 \theta$$ $$x \cos \theta + y \sin \theta = 5 (\sin^2 \theta + \cos^2 \theta) = 5$$

Thus, the equation of the tangent is:

$$x \cos \theta + y \sin \theta - 5 = 0$$

Step 5: Equation of the normal at $(5 \cos \theta, 5 \sin \theta)$
The slope of the normal is the negative reciprocal of the slope of the tangent:

$$\frac{\sin \theta}{\cos \theta}$$

Using the point-slope form of the normal:

$$y - 5 \sin \theta = \frac{\sin \theta}{\cos \theta}(x - 5 \cos \theta)$$

Simplifying:

$$y \cos \theta - 5 \sin \theta \cos \theta = x \sin \theta - 5 \sin \theta \cos \theta$$ $$x \sin \theta - y \cos \theta = 0$$

Thus, the equation of the normal is:

$$x \sin \theta - y \cos \theta = 0$$

ii. Ellipse $3x^2 + 2y^2 + 5x - 13y + 2 = 0$

Step 1: Differentiate implicitly to find the slope of the tangent
Differentiating the equation of the ellipse:

$$6x + 6y \frac{dy}{dx} + 5 - 13 \frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$:

$$\frac{dy}{dx} (6y - 13) = -6x - 5$$ $$\frac{dy}{dx} = \frac{6x + 5}{6y - 13}$$

Step 2: Calculate the slope at $\left(1, \frac{10}{3}\right)$
Substitute $x = 1$ and $y = \frac{10}{3}$ into the derivative:

$$\frac{dy}{dx} = \frac{6(1) + 5}{6\left(\frac{10}{3}\right) - 13} = \frac{11}{7}$$

Thus, the slope of the tangent at $\left(1, \frac{10}{3}\right)$ is $\frac{11}{7}$.

Step 3: Equation of the tangent at $\left(1, \frac{10}{3}\right)$
Using the point-slope form of the tangent:

$$y - \frac{10}{3} = -\frac{11}{7}(x - 1)$$

Simplifying:

$$7y - 70 = -11x + 11$$ $$11x + 7y - \frac{103}{3} = 0$$

Multiplying through by 3:

$$33x + 21y - 103 = 0$$

Thus, the equation of the tangent is $33x + 21y - 103 = 0$.

Step 4: Equation of the normal at $\left(1, \frac{10}{3}\right)$
The slope of the normal is the negative reciprocal of the slope of the tangent:

$$\frac{7}{11}$$

Using the point-slope form of the normal:

$$y - \frac{10}{3} = \frac{7}{11}(x - 1)$$

Simplifying:

$$11y - 110 = 7x - 7$$ $$7x - 11y + \frac{89}{3} = 0$$

Multiplying through by 3:

$$21x - 33y + 89 = 0$$

Thus, the equation of the normal is $21x - 33y + 89 = 0$.

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Key Formulas or Methods Used

• Point-slope form of the line: $y - y_1 = m(x - x_1)$
• Implicit differentiation: Used to find the slope of the tangent for both the circle and ellipse.
• Negative reciprocal: The slope of the normal line is the negative reciprocal of the tangent line’s slope.

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Summary of Steps

1. Differentiate the equation of the circle or ellipse to find the slope of the tangent.
2. Use the point-slope form to write the equation of the tangent.
3. Find the slope of the normal (negative reciprocal of the tangent’s slope).
4. Use the point-slope form again to find the equation of the normal.
5. Simplify the equations to the standard form.