Home
Class 12
MATHS
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 (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)x-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
Promotional Banner

Similar Questions

Explore conceptually related problems

The coordinates (2, 3) and (1, 5) are the foci of an ellipse which passes through the origin. Then the equation of the (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

Find the equation of tangent to the curve y=sin^(-1)(2x)/(1+x^2)a tx=sqrt(3)

Find the equation of the straight line which passes through the origin and makes angle 60^0 with the line x+sqrt(3)y+sqrt(3)=0 .

Solve : sqrt(3)x^(2)+5x+2sqrt(3)=0

Find the equation of tangent to y=int_(x^2)^(x^3)(dt)/(sqrt(1+t^2))a tx=1.

Find the equation of the tangent to the curve y= sqrt(3x-2) which is parallel to the line 4x-2y+5=0.

For the curve y = 4x^(3) – 2x^(5) , find all the points at which the tangent passes through the origin.

Radius of the circle that passes through the origin and touches the parabola y^2=4a x at the point (a ,2a) is (a) 5/(sqrt(2))a (b) 2sqrt(2)a (c) sqrt(5/2)a (d) 3/(sqrt(2))a

The equation of the lines passing through the point (1,0) and at a distance (sqrt(3))/2 from the origin is (a) sqrt(3)x+y-sqrt(3) =0 (b) x+sqrt(3)y-sqrt(3)=0 (c) sqrt(3)x-y-sqrt(3)=0 (d) x-sqrt(3)y-sqrt(3)=0