Question

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`

Answer 1

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