The given equation of the ellipse is (x²)/(16) + (y²)/(9) = 1. By comparing this with the standard form of an ellipse (x²)/(a²) + (y²)/(b²) = 1, we can identify the values of a² and b². The point through which the tangent passes is (8, 0).

The given equation of the ellipse is (x²)/(16) + (y²)/(9) = 1. By comparing this with the standard form of an ellipse (x²)/(a²) + (y²)/(b²) = 1, we can identify the values of a² and b². The point through which the tangent passes is (8, 0).

The general equation of a tangent to an ellipse is y = mx + c. For this line to be tangent to the ellipse (x²)/(a²) + (y²)/(b²) = 1, the condition c² = a²m² + b² must be satisfied. From the ellipse equation, we have a² = 16 and b² = 9.

The general equation of a tangent to an ellipse is y = mx + c. For this line to be tangent to the ellipse (x²)/(a²) + (y²)/(b²) = 1, the condition c² = a²m² + b² must be satisfied. From the ellipse equation, we have a² = 16 and b² = 9.

MODIFIED: Find the equation of the tangent to the ellipse variable_97²/16 + variable_97²/9 = 1 that passes through the point (8, 0).

Answer: The equations of the tangents are $y = \frac{\sqrt{3}}{4}x - 2\sqrt{3}$ and $y = -\frac{\sqrt{3}}{4}x + 2\sqrt{3}$ .

Step-by-step solution

Step 1: Identify the equation of the ellipse and the external point

The given equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$ . By comparing this with the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , we can identify the values of $a^2$ and $b^2$ . The point through which the tangent passes is $(8, 0)$ .

Step 2: Use the condition for tangency

The general equation of a tangent to an ellipse is $y = mx + c$ . For this line to be tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , the condition $c^2 = a^2m^2 + b^2$ must be satisfied. From the ellipse equation, we have $a^2 = 16$ and $b^2 = 9$ .

Step 3: Substitute values and the external point into the tangent equation

Substitute $a^2 = 16$ and $b^2 = 9$ into the tangency condition, giving $c^2 = 16m^2 + 9$ . Since the tangent passes through the point $(8, 0)$ , we substitute $x=8$ and $y=0$ into the tangent equation $y = mx + c$ , which gives $0 = m(8) + c$ , so $c = -8m$ .

Step 4: Solve for the slope 'm'

Now, substitute $c = -8m$ into the tangency condition $c^2 = 16m^2 + 9$ . This results in an equation solely in terms of $m$ . Solving this quadratic equation for $m$ will give us the slopes of the two possible tangent lines.

Step 5: Find the corresponding 'c' values

For each value of $m$ found, we can calculate the corresponding value of $c$ using the relation $c = -8m$ . This will give us two pairs of $(m, c)$ values, each corresponding to a unique tangent line.

Step 6: Write the equations of the tangent lines

Substitute the calculated values of $m$ and $c$ back into the general tangent equation $y = mx + c$ to obtain the equations of the two tangent lines.

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