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

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

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

To find the slope of the tangent line passing through ${\displaystyle (x_{0},y_{0})}$, use implicit differentiation. Differentiate both sides of the equation of the curve with respect to ${\displaystyle x}$. By the chain rule,

${\displaystyle 2x+4yy'=0.}$

Let ${\displaystyle y_{0}'}$ be the slope of the tangent line at point ${\displaystyle (x_{0},y_{0})}$. Assume ${\displaystyle y_{0}\neq 0}$. Then

${\displaystyle y'_{0}=-{\frac {x_{0}}{2y_{0}}}.}$

Using the point-slope formula, the tangent line equation to the curve at the point ${\displaystyle (x_{0},y_{0})}$ can be written as

${\displaystyle y-y_{0}=-{\frac {x_{0}}{2y_{0}}}(x-x_{0}).}$

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

${\displaystyle -6-y_{0}={\frac {x_{0}^{2}}{2y_{0}}}.}$

Together with Equation (1) we can solve for the point. ${\displaystyle (x_{0},y_{0})=\left({\frac {8}{3}},-{\frac {2}{3}}\right)}$ or ${\displaystyle \left(-{\frac {8}{3}},-{\frac {2}{3}}\right).}$

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

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