If $M_{1}$ and $M_{2}$ are the feet of the perpendiculars from the foci $S_{1}$ and $S_{2}$ of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ on the tangent at any point $P$ on the ellipse, then $\left( S_{1}M_{1} \right)(S_{2}M_{2})$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
We know that the product of perpendiculars drawn from two foci $S_{1}$ and $S_{2}$ of an ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ on the tangent at any point $P$ on the ellipse is equal to the square of the semiminor axis.
$	herefore\left( S_{1}M_{1} ight) \bullet \left( S_{2}M_{2} ight) = 16$

Q. If $M_{1}$ and $M_{2}$ are the feet of the perpendiculars from the foci $S_{1}$ and $S_{2}$ of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ on the tangent at any point $P$ on the ellipse, then $\left( S_{1}M_{1} \right)(S_{2}M_{2})$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?