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Question

The slope of normal to curve y=f(x), where x=t2−5t+4 and y=t2−7t+10 at (−2,−2), is

A
2
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B
1
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C
1
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D
12
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Solution

The correct option is C 1
Given curve : x=t25t+4
On differentiating both sides w.r.t. t
dxdt=2t5
and y=t27t+10
On differentiating on both sides w.r.t. t
dydt=2t7
For x=2
t25t+4=2
t25t+6=0
t=2,3
and For y=2
t27t+12=0
t=4,3
So, t=3 at point (2,2)
Now,
dydx=dydtdtdx=2t72t5
Slope of tangent at t=3
dydx(t=3)=1
slope of normal =1

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