JEE Advanced 2013 · Paper 2 · Mathematics Q57

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MathematicsList match+3 / -1Question 57 of 60

A line $L:y=mx+3$ meets $y$ -axis at $E(0,3)$ and the arc of the parabola $y^2=16x$ , $0\le y\le6$ 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 triangle $EFG$ has a local maximum.

Match List I with List II and select the correct answer using the code given below the lists:

List I

P
$m=$
Q
Maximum area of $\triangle EFG$ is
R
$y_0=$
S
$y_1=$

List II

1
$\frac12$
2
4
3
2
4
1

P → 4; Q → 1; R → 2; S → 3

P → 3; Q → 4; R → 1; S → 2

P → 1; Q → 3; R → 2; S → 4

P → 1; Q → 3; R → 4; S → 2

Answer and worked solution

Official answer: A

Set-up. $F=(4t^2,8t)$ with $0\le t\le\frac34$ . The tangent $ty=x+4t^2$ meets the y-axis at $G(0,4t)$ .

Area. Base $EG=3-4t$ on the y-axis, height $4t^2$ :
$A=\tfrac12(3-4t)(4t^2)=6t^2-8t^3,\qquad A'=12t-24t^2=0\implies t=\tfrac12.$

Values. $F=(1,4)$ : $y_0=4$ → 2; $y_1=2$ → 3; maximum area $\frac32-1=\frac12$ → 1; slope of EF $m=1$ → 4.

Code 4 1 2 3.

(A)

Topics:Maxima and minima Parabola

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JEE Advanced 2013 · Paper 2, Mathematics question 57