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  • Let C be a curve given by <img alt="y(x)=1+\sqrt{4x-3},x>\frac{3}{4}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D%20y%28x%29%3D1&plus;%5Csqrt%7B4x-3%7D%2Cx%3E%5Cfrac%

Let C be a curve given by y(x)=1+\sqrt{4x-3},x>\frac{3}{4}.

 If P is a point on C, such that the tangent at P has slope   \frac{2}{3}  ,

then a point through which the normal at P passes, is :

 

Option: 1

 (2, 3)


Option: 2

(4, −3)


Option: 3

 (1, 7)


Option: 4

 (3, −4)

 


Answers (1)

best_answer

Given curve is y(x)=1+\sqrt{4x-3},x>\frac{3}{4}

Let point P\;\text{ be }\;\left ( \alpha,1+\sqrt{4\alpha-3} \right )

\\\text{slope }=\frac{dy}{dx}=\frac{2}{\sqrt{4x-3}}

\\\frac{dy}{dx}\left . \right |_{at \;P}=\frac{2}{3}\\\frac{2}{\sqrt{4\alpha-3}}=\frac{2}{3}\\\alpha=3

Hence, point P (3,4)

\\\text{ slope of normal at }\mathrm{P\;}(3,4)\text{ is }=-\frac{3}{2}\\\text{ equation of normal }\\ Y-4=-\frac{3}{2}(X-3)\\ 2 y-8=-3 x+9 \\3 x+2 y=17\\ \text{ clearly it is passes through (1,7)}

Posted by

Suraj Bhandari

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