Question

Finding Tangents Find the equations of all lines tangent to \(y=9-x^{2}\) that pass through the point (1,12)

Short answer

The equations of the lines tangent to \(y=9-x^{2}\) that pass through the point (1,12) are \(y=0\) and \(y=16\).

Step-by-step solution

Compute the derivative

The derivative of \(y=9-x^{2}\) is \(y'= -2x\). This derivative gives the slope of the tangent line at any point \(x\).

Use point-slope form

The point-slope form of a line is given by \(y-y_{1}=m(x-x_{1})\), where \((x_{1}, y_{1})\) is a point on the line and \(m\) is the slope of the line. Since the line must pass through the point (1,12), substitute those values in to get \(y-12=-2x(x-1)\). This is the equation of all tangent lines, but we still have to solve for the specific \(x\) that makes the line touch the curve, i.e., the points of tangency.

Solve for x

Set \(y=9-x^2\) and \(y=12+2x(x-1)\) equal to each other to solve for \(x\), because where they intersect gives you the tangible points. This gives: \(9-x^2=12+2x(x-1)\), simplifying gives \(0=3+x^2-2x\) or \(x^2-2x-3=0\). Factoring this gives \((x-3)(x+1)=0\), so \(x = 3\) or \(x = -1\). Therefore, the points of tangency are at \(x = 3\) and \(x = -1\).

Find the equations of the tangent lines

Substitute \(x=3\) and \(x=-1\) into the general line equation from Step 2 to get the equations of the two tangent lines. For \(x=3\), \(y-12=-2*3(3-1)= -12\) simplifies to \(y=-12+12=0\). So, one tangent line is \(y=0\). For \(x=-1\), \(y-12=-2*(-1)(-1-1) = 4\) simplifies to \(y=4+12=16\). So, the other tangent line is \(y=16\).