Chapter 9.1, Problem 94E

To determine

Tofind:The equation of the tangent line to the parabola $x^2=2y$ at the point $\left(-3,\frac{9}{2}\right)$ .

Answer to Problem 94E

The equation of the tangent line to the parabola $x^2=2y$ at point $\left(-3,\frac{9}{2}\right)$ is $6x+2y+9=0$ .

Explanation of Solution

Given information:

The equation of parabola is $x^2=2y$ and the point of tangencyis $\left(-3,\frac{9}{2}\right)$ .

Calculation:

Compare the equation of parabola with general equation of parabola ${\left(x-h\right)}^2=4p\left(y-k\right)$ .

This gives the values $h=0$ , $k=0$ and $p=\frac{1}{2}$ .

So, the vertex of the parabola is $\left(h,k\right)=\left(0,0\right)$ and the focus of the parabola is $\left(h,k+p\right)=\left(0,\frac{1}{2}\right)$ .

Now to find the slope of the tangent suppose that y-intercept of the tangent line is $\left(0,b\right)$ .

Then the distance between the distance between the focus and y-intercept is equal to the distance between the point of tangency and focus.

Firstly, calculate the distance between the focus and y-intercept.

  $\begin{array}{c}{d}_1=\sqrt{{\left(0-0\right)}^2+{\left(\frac{1}{2}-b\right)}^2}\\ =\frac{1}{2}-b\end{array}$

Now calculate the distance between the point of tangency and focus.

  $\begin{array}{c}{d}_2=\sqrt{{\left(-3-0\right)}^2+{\left(\frac{9}{2}-\frac{1}{2}\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

Equate both the distance as they are equal.

  $\begin{array}{c}{d}_1=d_2\\ \frac{1}{2}-b=5\\ b=\frac{1}{2}-5\\ b=-\frac{9}{2}\end{array}$

Now the slope of the tangent line.

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{\frac{9}{2}-\left(-\frac{9}{2}\right)}{-3-0}\\ =\frac{9}{-3}\\ =-3\end{array}$

So, the equation of the tangent line:

  $\begin{array}{c}y=mx+b\\ y=-3x-\frac{9}{2}\\ 6x+2y+9=0\end{array}$

Therefore, the equation of the tangent line to the parabola $x^2=2y$ at point $\left(-3,\frac{9}{2}\right)$ is $6x+2y+9=0$ .