Question
A point on the ellipse $${x^2} + 3{y^2} = 9,$$ where the tangent is parallel to the line $$y - x = 0,$$ is :
A.
$$\left( {\sqrt 3 ,\,\sqrt 2 } \right)$$
B.
$$\left( { - \frac{{3\sqrt 3 }}{2},\, - \frac{{\sqrt 3 }}{2}} \right)$$
C.
$$\left( { - \frac{{3\sqrt 3 }}{2},\,\frac{{\sqrt 3 }}{2}} \right)$$
D.
$$\left( { - \sqrt 3 ,\,\sqrt 2 } \right)$$
Answer :
$$\left( { - \frac{{3\sqrt 3 }}{2},\,\frac{{\sqrt 3 }}{2}} \right)$$
Solution :
The point is $$\left( {{x_1},\,{y_1}} \right)$$ if $$x{x_1} + 3y{y_1} = 9$$ has the slope $$1,$$ i.e., $$ - \frac{{{x_1}}}{{3{y_1}}} = 1$$ and $$x_1^2 + 3y_1^2 = 9.$$ Solve the two equations.