Question

 Show that the line x + y − 2 = 0 is a tangent to the parabola x²+ 8y = 0 and find the position vector of the point of contact.

Answer 1

$$x^2+8y=0$$

$$y=-\dfrac{1}{8}x^2$$

$$y'=-\dfrac{1}{4}x$$

$$y-y_1=y'(x-x_1)$$

$$y=-\dfrac{1}{4}x_1(x-x_1)-\dfrac{1}{8}x_1^2$$

$$y=-\dfrac{1}{8}x_1x+\dfrac{1}{4}x_1^2$$

$$x + y − 2 = 0=>y=-x+2$$

$$\begin{matrix} -\dfrac{1}{4}x_1=-1 \\ \\ \dfrac{1}{8}x_1^2=2 \end{matrix}$$

The system has the solution $x_1=4.$

$$y_1=-\dfrac{1}{8}(4)^2=-2$$

The line $x+y-2=0$ is a tangent to the parabola $x^2+8y=0$ at the point $(4, -2).$

The position vector of the point of contact is $\langle4, -2\rangle.$