Question 144 of 212MediumMCQ
Which of the following points lies on the
tangent to the curve
x4ey + 2\[\sqrt {y + 1} \] = 3 at the
point (1, 0)?
Solutions · 1
Official solution
x4ey + 2\[\sqrt {y + 1} \] = 3
Differentiating w.r.t. x, we get
x4eyy' + ey4x3 + \[{{2y'} \over {2\sqrt {y + 1} }}\] = 0
At P(1, 0)
yP′ + 4 + yP′ = 0
\[ \Rightarrow \] yP′ = -2
Tangent at P(1, 0) is
y -- 0 = -- 2 (x -- 1)
2x + y = 2
Only (--2, 6) lies on it. Answer the question to unlock