# Perpendiculars from foci on any tangent of ellipse

Let Q and Q' be the feet of perpendiculars from foci S and S' to the tangent at a point P on an ellipse with eccentricity $$=\frac12$$. Given that SQ$$=2$$S'Q' and S'P=4. If SP and S'Q' intersect at R then find the lengths of SP, SQ, SR and QQ'.

My Attempt:

We know that the product of the lengths of perpendiculars drawn foci on any tangent of ellipse is equal to the square of its semi-minor axis.

Let S'Q'$$=x$$ and semi-minor axis$$=b$$

So, $$2x^2=b^2$$

Also, sum of focal distances of a point is equal to the length of major axis.

Let semi-major axis$$=a$$

So, $$SP+4=2a$$

Not able to join these dots.

• I think that $SQ=2\color{red}{S}Q'$ is impossible since $\triangle{SQQ'}$ is a right triangle with $\angle{SQQ'}=90^\circ$. I think it should be $SQ=2\color{red}{S'}Q'$. Commented Apr 10, 2023 at 11:27
• @mathlove yes, you are right. Sorry for the typo Commented Apr 10, 2023 at 15:13

We have $$\angle{SPQ}=\angle{S'PQ'}$$. (see here)

Since $$\triangle{QSP}$$ and $$\triangle{Q'S'P}$$ are similar, we get $$SQ:S'Q'=SP:S'P$$, and so $$SP=\dfrac{SQ}{S'Q'}\times S'P=2\times 4=8$$

As you wrote, we have $$SP+4=2a$$ from which $$a=6$$ follows.

Since the eccentricity is $$\dfrac 12$$, we have $$\dfrac{1}{2}=\dfrac{\sqrt{a^2-b^2}}{a}$$. So we get $$b=3\sqrt 3$$.

Also, as you wrote, we have $$2S'Q'^2=b^2$$, so we get $$S'Q'=\dfrac{3\sqrt 6}{2}$$ and $$SQ=2S'Q'=3\sqrt 6$$.

Since $$\triangle{S'Q'P}\equiv\triangle{RQ'P}$$, we have $$PR=PS'$$, so $$SR=PR+SP=4+8=12$$.

Finally, we have $$QQ'=PQ'+PQ=3PQ'=3\sqrt{S'P^2-S'Q'^2}=\dfrac{3\sqrt{10}}{2}$$