Question

The points on the ellipse $x^2+4y^2=2$, where the tangents are parallel to the line $x-2y-6=0$ are:

• A. $(1,-\frac{1}{2}),(-1,\frac{1}{2})$
• B. $\left(\frac{1}{2},-1\right),(\frac{1}{2},1)$
• C. $(-1,\frac{1}{2}),(1,\frac{1}{2})$
• D. $(-1,-1),(1,1)$

Answer 1

The correct option is A $(1,-\frac{1}{2}),(-1,\frac{1}{2})$

The given ellipse is $x^2+4y^2=2$

The tangents on this ellipse are parallel to the line $x-2y-6=0\Rightarrow x-2y=6$

So the slope of the line and the tangents should be equal.

So the slope of line $x-2y=6\Rightarrow y=\frac{x}{2}-3$ is $m=\frac{1}{2}$

Now ellipse$x^2+4y^2=2$

Differentiate wrt x on both sides to get $y^{{}^{\prime }}=\frac{-x}{4y}$

Now $y^{{}^{\prime }}=\frac{-x}{4y}=\frac{1}{2}\Rightarrow x=-2y$

we check the given option points for this relation and see that option A $(1,-\frac{1}{2}),(-1,\frac{1}{2})$ satisfy the relation.