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

Let $(x_{0},y_{0})$ be a point on the curve. Then

$x_{0}^{2}+2y_{0}^{2}=8.~~~~(1)$ 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,

$2x+4yy'=0.$ Let $y_{0}'$ be the slope of the tangent line at point $(x_{0},y_{0})$ . Assume $y_{0}\neq 0$ . Then

$y'_{0}=-{\frac {x_{0}}{2y_{0}}}.$ 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. $(x_{0},y_{0})=\left({\frac {8}{3}},-{\frac {2}{3}}\right)$ or $\left(-{\frac {8}{3}},-{\frac {2}{3}}\right).$ 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 $\left({\frac {8}{3}},-{\frac {2}{3}}\right),~\left(-{\frac {8}{3}},-{\frac {2}{3}}\right)$ .