The point which lies on the tangent drawn to | vedclass

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Question

(D) Given curve: $x^4 e^y + 2 \sqrt{y+1} = 3$. Differentiating both sides with respect to $x$: $x^4 e^y y' + 4x^3 e^y + \frac{2 y'}{2 \sqrt{y+1}} = 0$

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$(-2,6)$

Vedclass · Answer

(D) Given curve: $x^4 e^y + 2 \sqrt{y+1} = 3$. Differentiating both sides with respect to $x$: $x^4 e^y y' + 4x^3 e^y + \frac{2 y'}{2 \sqrt{y+1}} = 0$. At the point $(1,0)$,substitute $x=1$ and $y=0$: $(1)^4 e^0 y' + 4(1)^3 e^0 + \frac{y'}{\sqrt{0+1}} = 0$. $y' + 4 + y' = 0 \Rightarrow 2y' = -4 \Rightarrow y' = -2$. The equation of the tangent at $(1,0)$ with slope $m = -2$ is: $y - 0 = -2(x - 1) \Rightarrow y = -2x + 2 \Rightarrow 2x + y = 2$. Now,check which point satisfies the equation $2x + y = 2$: For option $D$ $(-2, 6)$: $2(-2) + 6 = -4 + 6 = 2$. Thus,the point $(-2, 6)$ lies on the tangent.

The point which lies on the tangent drawn to the curve $x^4 e^y + 2 \sqrt{y+1} = 3$ at the point $(1,0)$ is

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