Science:Math Exam Resources/Courses/MATH100/December 2018/Question 04/Solution 1

Let $(x_0,y_0)$ be a point on the curve. Then this point satisfies

$x_0^2+2y_0^2=8.(1)$

To find the slope of the tangent line passing through $(x_0,y_0)$, use implicit differentiation. Differentiate both sides of the equation of the curve with respect to $x$. By the chain rule,

$2x+4y{y}^{\prime }=0.$

Let ${y_0}^{\prime }$ be the slope of the tangent line at point $(x_0,y_0)$. Assume $y_0\ne 0$. Then

$y{\text{'}}_0=-\frac{x_0}{2y_0}.$

Using the point-slope formula, the tangent line equation to the curve at the point $(x_0,y_0)$ can be written as

$y-y_0=-\frac{x_0}{2y_0}(x-x_0).$

Let this tangent line pass through $(0,-6)$. Putting $x=0$ and $y=-6$,

$-6-y_0=\frac{x_0^2}{2y_0}.$

Together with Equation (1) we can solve for the point. $(x_0,y_0)=(\frac{8}{3},-\frac{2}{3})$ or $(-\frac{8}{3},-\frac{2}{3}).$

Now consider if $y_0=0$, then $x_0=\pm 2\sqrt{2}.$ The slope of the tangent line DNE. It follows that the tangent line is a vertical line, either $x=2\sqrt{2}$ or $x=-2\sqrt{2}$. Neither of them passes through the point $(0,-6).$

Therefore, the points on the curve that satisfy the requirements are $(\frac{8}{3},-\frac{2}{3}),(-\frac{8}{3},-\frac{2}{3})$.