    Question

# Equation of tangent at that point of the curve y=1−ex2, where it meets y-axis

A
x+2y=0
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B
2x+y=0
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C
xy=2
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D
None of these
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Solution

## The correct option is C x+2y=0 Given , y=1−ex2 ……(1) Differentiate w.r. t x both side we get dydx=−ex2×12 …..(2) Given that graph ofequation (1) meets y-axis ⇒x=0 Put in equation (1), we get y=1−e02 y=1−e0 y=1−1 y=0 Put y=0 in equation (1) we getx=0 Now, put the value of x,y in equation (2), we get (dydx)(0,0)=−e02×12=−1×12=−12 Now required equation is y−y1=dydx(x−x1) y−0=−12(x−0) y=−12x 2y+x=0This is the required equation of tangent.  Suggest Corrections  0      Similar questions  Related Videos   Geometrical Interpretation of a Derivative
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