Equation of the normal to the curve y = andndash;+ 2 at the point of its intersection with the curve y = tan (tanandndash;1x) isa)2x andndash;y andndash;1 = 0b)2x andndash;y + 1 = 0c)2x + y andndash;3 = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer?

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Equation of the normal to the curve y = –

+ 2 at the point of its intersection with the curve y = tan (tan–¹ x) is

a)
2x – y – 1 = 0
b)
2x – y + 1 = 0
c)
2x + y – 3 = 0
d)
None of these

Correct answer is option 'A'. Can you explain this answer?

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Equation of the normal to the curve y = –+ 2 at the point of its...

y= −(x)^1/2+2....(i)
And bisector of first quadrant is y=x ..... (ii)
On solving Eqs(i) and (ii) we get
x=1,4
∴ From Eq (i)
Points are (1,1) and (4,0)
But (4,0) not satisfy Eq (ii)
∴ Point (1,1) is only point of intersection of curve (i) and line (ii)
Now, slope of tangent of curve (i) is dy/dx = -1/(2(x)1/2)
∴ Slope of tangent at point (1,1) is −1/|dy/dx|(1,1) = -1/2
−1/|dy/dx|(1,1) = -1/(−1/2)
= 2
∴ Equation of the normal to the curve at point (1,1) will be
y−1=2(x−1)
⇒2x−y−1=0

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Equation of the normal to the curve y = –+ 2 at the point of its intersection with the curve y = tan (tan–1x) isa)2x –y –1 = 0b)2x –y + 1 = 0c)2x + y –3 = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer?

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