Question

Q is a point on the auxiliary circle of an ellipse. P is the corresponding point on ellipse. N is the foot of perpendicular from focus S, to the tangent of auxillary circle at Q. Then

• A. SP = SN
• B. SP = PQ
• C. PN = SP
• D. NQ = SP

Answer 1

The correct option is A

SP = SN

$P=(a\text{cos}\theta ,b\text{sin}\theta )$

$Q=(a\text{cos}\theta ,a\text{sin}\theta )$

Tangent at Q is $(y-a\text{sin}\theta )=-\frac{1}{\text{tan}\theta }(x-a\text{cos}\theta )$

$x+y\text{tan}\theta -\frac{a}{\text{cos}\theta }=0$ and $(ae,0)$

$\therefore \text{SN}=\frac{ae-\frac{a}{\text{cos}\theta }}{\sqrt{1+{\text{tan}}^2\theta }}=|\text{ae cos}\theta -\text{a}|$

$\text{SP}=\sqrt{(ae-a\text{cos}\theta {)}^2+b^2{\text{sin}}^2\theta }$

$=\sqrt{a^2{e}^2+a^2{\cos }^2\theta -2a^{2}e\cos \theta +a^{2}si{n}^2\theta -a^2{e}^2{\sin }^2\theta }$

$=\sqrt{a^2{e}^2{\cos }^2\theta -2a^{2}e\cos \theta +a^2}=|\text{ae cos}\theta -\text{a}|$

$\therefore$ SP = SN