6.2 Q-2

Question Statement

The problem asks to find the equations of the tangent and normal lines to the circle given by the equation $4x^2 + 4y^2 - 16x + 24y - 117 = 0$ at the points on the circle where the abscissa (x-coordinate) is $x = -4$.

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Background and Explanation

Before solving the problem, let’s review the concepts needed:

• Tangent to a Circle: The equation of the tangent at a point on a circle can be derived using the derivative of the circle’s equation. Alternatively, for a point $(x_1, y_1)$ on the circle, the tangent equation is given by:

$$T = x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0$$

where $g$, $f$, and $c$ are constants from the general equation of the circle.

• Normal to a Circle: The normal at a point is a line that is perpendicular to the tangent at that point. The equation of the normal can be derived from the slope of the tangent.

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Solution

Step 1: Simplify the Circle Equation

The given equation of the circle is:

$$4x^2 + 4y^2 - 16x + 24y - 117 = 0$$

To simplify, divide the entire equation by 4:

$$x^2 + y^2 - 4x + 6y - \frac{117}{4} = 0$$

Step 2: Find the Points on the Circle Where $x = -4$

Substitute $x = -4$ into the simplified equation:

$$(-4)^2 + y^2 - 4(-4) + 6y - \frac{117}{4} = 0$$

Simplifying the terms:

$$16 + y^2 + 16 + 6y - \frac{117}{4} = 0$$ $$y^2 + 32 + 6y - \frac{117}{4} = 0$$

Multiply everything by 4 to eliminate the fraction:

$$4y^2 + 24y + 128 - 117 = 0$$ $$4y^2 + 24y + 11 = 0$$

Now solve the quadratic equation $4y^2 + 24y + 11 = 0$ using factoring or the quadratic formula.

Factoring:

$$(2y + 1)(2y + 11) = 0$$

So the solutions are:

$$y = -\frac{1}{2} \quad \text{or} \quad y = -\frac{11}{2}$$

Thus, the points on the circle where $x = -4$ are $\left(-4, -\frac{11}{2}\right)$ and $\left(-4, -\frac{1}{2}\right)$.

Step 3: Equation of the Tangent at $\left(-4, -\frac{11}{2}\right)$

For the point $\left(-4, -\frac{11}{2}\right)$, the equation of the tangent is:

$$-4x + \left(\frac{-11}{2}\right)y + (-2)(x + (-4)) + 3\left(y + \left(\frac{-11}{2}\right)\right) - \frac{117}{4} = 0$$

Simplifying:

$$-4x + \left(\frac{-11}{2}\right)y + 2x + 8 + 3y - \frac{33}{2} - \frac{117}{4} = 0$$

Combine like terms:

$$-6x + 3y - \frac{11}{2}y + 8 - \frac{33}{2} - \frac{117}{4} = 0$$

Multiply through by 4 to clear the denominators:

$$24x + 12y - 22y + 32 - 66 - 117 = 0$$

Simplifying:

$$24x - 10y - 151 = 0$$

Thus, the equation of the tangent at $\left(-4, -\frac{11}{2}\right)$ is:

$$24x - 10y + 151 = 0$$

Step 4: Equation of the Tangent at $\left(-4, -\frac{1}{2}\right)$

For the point $\left(-4, -\frac{1}{2}\right)$, the equation of the tangent is:

$$-4x + \left(\frac{-1}{2}\right)y + (-2)(x + 4) + 3\left(y + \left(\frac{-1}{2}\right)\right) - \frac{117}{4} = 0$$

Simplifying:

$$-4x + \left(\frac{-1}{2}\right)y - 2x + 8 + 3y - \frac{117}{4} = 0$$

Multiply through by 4 to clear the denominator:

$$-24x + 10y - 91 = 0$$

Thus, the equation of the tangent at $\left(-4, -\frac{1}{2}\right)$ is:

$$24x - 10y + 91 = 0$$

Step 5: Equation of the Normal at $\left(-4, -\frac{11}{2}\right)$

The equation of the normal can be found using the point-slope form. The slope of the normal is $\frac{12}{5}$. Using the point-slope form:

$$y + \frac{11}{2} = \frac{12}{5}(x + 4)$$

Simplifying:

$$12y + 66 = 5x + 20$$

Thus, the equation of the normal at $\left(-4, -\frac{11}{2}\right)$ is:

$$15x - 12y - 46 = 0$$

Step 6: Equation of the Normal at $\left(-4, -\frac{1}{2}\right)$

Similarly, the equation of the normal at $\left(-4, -\frac{1}{2}\right)$ can be derived. The slope of the normal is $-\frac{1}{2}$. The equation of the normal is:

$$y + \frac{1}{2} = \frac{5}{2}(x + 4)$$

Simplifying:

$$12y - 6 = 5x + 20$$

Thus, the equation of the normal at $\left(-4, -\frac{1}{2}\right)$ is:

$$5x + 12y + 26 = 0$$

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Key Formulas or Methods Used

• Point-Slope Form of a Line: $y - y_1 = m(x - x_1)$
• General Equation of the Tangent: $T = x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0$
• Implicit Substitution: Substituting the given value of $x$ into the equation of the circle to find corresponding $y$-coordinates.

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Summary of Steps

1. Simplify the given equation of the circle.
2. Substitute $x = -4$ into the simplified equation to find the corresponding $y$-coordinates.
3. Use the point-slope form to find the equation of the tangent at each point.
4. Use the negative reciprocal of the tangent’s slope to find the equation of the normal.
5. Simplify the equations to their standard forms.