If P_1 and P_2 are two points on the ellipse x^24+y^2=1 at which the tangents are parallel…

Question

If $P_1$ and $P_2$ are two points on the ellipse $\dfrac{x^2}{4}+y^2=1$ at which the tangents are parallel to the chord joining the points $(0,1)$ and $(2,0)$, then the distance between $P_1$ and $P_2$ is:

Accepted Answer

Slope of chord $=\dfrac{1-0}{0-2}=-\tfrac12$. For $a=2,b=1$, points of tangency for slope $m$: $\left(\mp\dfrac{a^2m}{\sqrt{a^2m^2+b^2}},\pm\dfrac{b^2}{\sqrt{a^2m^2+b^2}}\right)$.

With $m=-\tfrac12$: $a^2m^2+b^2=4\cdot\tfrac14+1=2$, $\sqrt2$.
$$P_{1,2}=\left(\pm\sqrt2,\ \mp\frac{1}{\sqrt2}\right).$$
Distance $=\sqrt{(2\sqrt2)^2+\left(\dfrac{2}{\sqrt2}\right)^2}=\sqrt{8+2}=\sqrt{10}.$

Solutions · 1