1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $SQ=2\color{red}{S'}Q'$. $\endgroup$
    – mathlove
    Commented Apr 10, 2023 at 11:27
  • $\begingroup$ @mathlove yes, you are right. Sorry for the typo $\endgroup$
    – aarbee
    Commented Apr 10, 2023 at 15:13

1 Answer 1

2
+50
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .