MODIFIED: Find the equation of the tangent to the ellipse variable_97²/16 + variable_97²/9 = 1 that passes through the point (8, 0).
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Step-by-Step Solution
Step 1: Identify the equation of the ellipse and the external point
The given equation of the ellipse is 16x2+9y2=1. By comparing this with the standard form of an ellipse a2x2+b2y2=1, we can identify the values of a2 and b2. 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 a2x2+b2y2=1, the condition c2=a2m2+b2 must be satisfied. From the ellipse equation, we have a2=16 and b2=9.
Step 3: Substitute values and the external point into the tangent equation
Substitute a2=16 and b2=9 into the tangency condition, giving c2=16m2+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 c2=16m2+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.