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The coordinates (2, 3) and (1, 5) are th...

The coordinates (2, 3) and (1, 5) are the foci of an ellipse which passes through the origin. Then the equation of the tangent at the origin is `(3sqrt(2)-5)x+(1-2sqrt(2))y=0` tangent at the origin is `(3sqrt(2)+5)x+(1+2sqrt(2)y)=0` tangent at the origin is `(3sqrt(2)+5)x-(2sqrt(2+1))y=0` tangent at the origin is `(3sqrt(2)-5)-y(1-2sqrt(2))=0`

A

tangent at the origin is `(3sqrt(2)-5)x+(1-2sqrt(2))y=0`

B

tangent at the origin is `(3sqrt(2)-5)x+(1+2sqrt(2))y=0`

C

tangent at the origin is `(3sqrt(2)-5)x-(2sqrt(2)+1)y=0`

D

tangent at the origin is `(3sqrt(2)-5)-y-(1-2sqrt(2))=0`

Text Solution

Verified by Experts

Tangent and normal are bisector of `angleSPS`
Now, the eqaution of SP is `y=3x//2` and that of SP is y=5,
Then equation of angle bisectors are
`(3x-2y)/(sqrt(13))=+-(5x-y)/(sqrt(26))`
or `3x-2y=+(5x-y)/(sqrt(2))`
Therefore ,the lines are
`(3sqrt(2)-5)x+(1-2sqrt(2))y=0`
and `(3sqrt(2)+5)x-2sqrt(2)+1)y=0`
Now, (2,3) and (1,5) line on the same side of `(3sqrt(3)-5)x+(1-2sqrt(2)y=0`, which is the equation of tangent.

Points (2,3) adn (1,5) a line on the different sides of `(3sqrt(2)+5)x+(2-sqrt(2)+1)y=0`, which is the equation of normal
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