Chapter 9, Problem 23RE

To determine

Calculate the equation of tangent line to the parabola.

Answer to Problem 23RE

The equation of the tangent is $y=-2x+2$ .

Explanation of Solution

Given: It is given in the question that the equation is $x^2=-2y$ and the point is $(2,-2)$ .

Concept Used:

In this, use the concept of parabola and also make the equation in standard form and from that find the vertex and focus.

Calculation: Here, the equation is $x^2=-2y$ .

Rewrite the equation of the parabola in standard from,

  $\begin{array}{l}{x}^2=-2y\\ x^2=4(-\frac{1}{2})y\end{array}$

Now, find the vertex and focus from the above standard form: $Vertex:(0,0);Focus=(0,-\frac{1}{2})$ .

Now, write the equation for $d_1$ , the distance between the focus and the desired y − intersept.

  $d_1=\frac{1}{2}+b$

Also, find $d_2$ , the distance between the given point and the focus.

  $\begin{array}{l}{d}_2=\sqrt{{(2-0)}^2+{(-2-\frac{-1}{2})}^2}\\ d_2=\sqrt{4+{(-\frac{3}{2})}^2}\\ d_2=\sqrt{4+\frac{9}{4}}=\frac{5}{2}\end{array}$

Now, equate $d_1$ and $d_2$ and solve for b.

  $\frac{1}{2}+b=\frac{5}{2}$

  $b=\frac{5}{2}-\frac{1}{2}=2$

Now, find the slope of the line by using b.

  $m=\frac{-2-(2)}{2-0}=\frac{-4}{2}=-2$

Now, make the equation:

  $y=-2x+2$

Now, to find x − intercept of the line, let $y=0$ and solve for x :

  $\begin{array}{l}-2x+2=0\\ -2x=-2\\ x=1\end{array}$

Conclusion:

The equation is $y=-2x+2$ .