Home
Class 12
MATHS
A line L:y=mx+3 meets y-axis at E(0,3) a...

A line L:y=mx+3 meets y-axis at E(0,3) and the arc of the parabola y 2 =16x,0≤y≤6 at the point F(x 0 ​ ,y 0 ​ ). The tangent to the parabola at F(x 0 ​ ,y 0 ​ ) intersects the y-axis at G(0,y 1 ​ ). The slope m of the line L is chosen such that the area of the △EFG has a local maximum. Match List 1 with List 2 List 1 List 2 A. m= 1. 2 1 ​ B. Maximum area of ΔEFG is 2. 4 C. y 0 ​ = 3. 2 D. y 1 ​ = 4. 1

A

`{:(P,, Q,, R,, S), (4,, 1,, 2,, 3):}`

B

`{:(P,, Q,, R,, S), (3,, 4,, 1,, 2):}`

C

`{:(P,, Q,, R,, S), (1,, 3,, 2,, 4):}`

D

`{:(P,, Q,, R,, S), (1,, 3,, 4,, 2):}`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Here, `y ^(2) = 16x, 0 le y le 6`

Tangent at `F, y t = x + at ^(2)`
At `" " x = 0, y = at = 4t`
Also, ` ( 4 t ^(2), 8 t) ` satisfy `y = mx + c `
`rArr " " 8t k= 4m t ^(2) + 3 `
`rArr " " 4 mt^(2) - 8 t + 3 = 0`
`therefore" "` Area of ` Delta = (1)/(2) |{:( 0,, 3,,1 ), ( 0,, 4t,, 1), (4t^(2),, 8t,, 1):}| = (1)/(2) * 4t^(2)( 3- 4t)`
`" " A = 2 [ 3t^(2) - 4t^(3) ]`
`therefore " " ( dA)/( dt) = 2 [ 6t - 12 t ^(2)] = - 12 t (12 t - 1)`
` (##41Y_SP_MATH_C10_E04_038_S02##) `
`therefore ` Maximum at ` t = (1)/(2) and 4 mt ^(2) - 8t + 3 =0`
`rArr m - 4 + 3 = 0`
`rArr " " m = 1 `
`" " G(0, 4t) rArr G (0, 2)`
`rArr " " y _1= 2 `
`(x _0, y _0) = ( 4t^(2) , 8t) = ( 1, 4)`
`" " y _0 = 4 `
`therefore" " ` Area` = 2((3)/(4) - (1)/(2)) = (1)/(2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

Axis of the parabola x^2 - 3y - 6x + 6 = 0 is

Tangents are drawn to the parabola y^2=4a x at the point where the line l x+m y+n=0 meets this parabola. Find the point of intersection of these tangents.

A tangent to the parabola y^2=8x makes an angle of 45^0 with the straight line y=3x+5. Then find one of the points of contact.

The area of the region bounded by the parabola (y-2)^(2) = x- 1 , the tangent to the parabola at the point (2,3) and the x-axis is

If y=m_1x+c and y=m_2x+c are two tangents to the parabola y^2+4a(x+a)=0 , then

A movable parabola touches x-axis and y-axis at (0,1) and (1,0). Then the locus of the focus of the parabola is :

The point of contact of the tangent 2x+3y+9=0 to the parabola y^(2)=8x is:

The point of contact of the line 2x-y+2=0 with the parabola y^(2)=16x is:

Consider the parabola whose focus is at (0,0) and tangent at vertex is x-y+1=0 Tangents drawn to the parabola at the extremities of the chord 3x+2y=0 intersect at angle