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Let d1a n dd2 be the length of the perpe...

Let `d_1a n dd_2` be the length of the perpendiculars drawn from the foci `Sa n dS '` of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` to the tangent at any point `P` on the ellipse. Then, `S P : S^(prime)P=` (a)`d_1: d_2` (b) `d_2: d_1` (c)`d_1 ^2:d_2 ^2` (d) `sqrt(d_1):sqrt(d_2)`

A

`d_(1):d_(2)`

B

`d_(2):d_(1)`

C

`d_(1)^(2):d_(2)^(2)`

D

`sqrt(d_(1)):sqrt(d_(1))`

Text Solution

Verified by Experts

Tangent at `P( a cos alpha, b sin alpha)` is
`(x)/(a)cos alpha+(y)/(b) sin alpha=1" "(1)`
The distances of focus, S(ae,0) from this tangent is
`d_(1)=(|e cos alpha-1|)/(sqrt((cos^(2))/(a^(2))+(sin^(2)alpha)/(b^(2))))`
`d_(1)=(cos alpha)/(sqrt((cos^(2))/(a^(2))+(sin^(2)alpha)/(b^(2))))`
The distance of fous S'(ae,0) from this line is
`d_(2)=(1+ecos alpha)/(sqrt((cos^(2)alpha)/(a^(2))+(sin^(2)alpha)/(b^(2))))`
`(d_(1))/(d_(2))=(1-ecos alpha)/(1+ecosalpha)`
Now, SP = `a-ae cos alpha and S'P=a+ae cos alpha`
or `(SP)/(S'P)=(1-e cos alpha)/(1+ecos alpha)=(d_(1))/(d_(2))`
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