Let the tangents at the points P and Q on the ellipse x^22+y^24=1 meet at the point R(√2,2…

Question

Let the tangents at the points $P$ and $Q$ on the ellipse $\dfrac{x^2}2+\dfrac{y^2}4=1$ meet at the point $R(\sqrt2,2\sqrt2-2)$. If $S$ is the focus of the ellipse on its negative major axis, then $SP^2+SQ^2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 13

Accepted Answer

Major axis is along $y$ ($b^2=4>a^2=2$), $c^2=4-2=2$, so $S=(0,-\sqrt2)$.

Tangent at $(x_1,y_1)$: $\dfrac{xx_1}2+\dfrac{yy_1}4=1$. Through $R$: $\sqrt2x_1+(\sqrt2-1)y_1=2$ --- this is the chord of contact $PQ$.

Solving with $2x^2+y^2=4$ gives $P=(\sqrt2,0)$ and $Q=(1,\sqrt2)$ (both verified on the ellipse and on the chord).

$$SP^2=(\sqrt2)^2+(\sqrt2)^2=4,\qquad SQ^2=1^2+(2\sqrt2)^2=1+8=9.$$

$SP^2+SQ^2=4+9=13.$

Solutions · 1