NARAYANA IIT/PMT ACADEMY - Narayanaicc.com
NARAYANA IIT/PMT ACADEMY - Narayanaicc.com
NARAYANA IIT/PMT ACADEMY - Narayanaicc.com
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<strong>NARAYANA</strong> <strong>IIT</strong>/<strong>PMT</strong> <strong>ACADEMY</strong><br />
JEE ADVANCE : 2013<br />
(PAPER – II) CODE: 0<br />
(0, 3)<br />
(0, 3)<br />
E<br />
G<br />
0, 0<br />
F(x , y )<br />
0 0<br />
(0, y ) 1<br />
2<br />
y =16x<br />
Let Co-ordinate of F= (x 0 , y 0 ) = (4t 2 8t)<br />
Co-ordinate of E = (0, 3) ……… given<br />
Equation of tangent at F ty = x + 4t 2<br />
Co-ordinate of G (0, 4t)<br />
0 3 1<br />
Area of EFG<br />
1 0 4 t 1<br />
2 4<br />
2<br />
t 8 t 1<br />
1 12 2 3<br />
t t<br />
2<br />
Let<br />
t<br />
1 12<br />
2 16<br />
3<br />
t t<br />
2<br />
' t<br />
1<br />
2<br />
24t 48t<br />
2<br />
' t 0 t<br />
1<br />
2<br />
" t<br />
1<br />
24<br />
2<br />
96t<br />
"<br />
1 1 1<br />
24 96<br />
2 2 2<br />
12<br />
Since (4t 2 , 8t) lies on the lie y = mx + 3<br />
8t<br />
3<br />
m<br />
2<br />
4t<br />
1<br />
at t , t will be maximum<br />
2<br />
2<br />
8t 4mt<br />
3<br />
At t = 1 2<br />
We get 1 m<br />
Maximum area will be at t = 1/2<br />
maximum area of triangle = 1/2<br />
y 0 = 8t = 4<br />
y 1 = 4t = 2<br />
<strong>NARAYANA</strong> GROUP OF EDUCATIONAL INSTITUTIONS - INDIA<br />
39
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