A) (3, - 4)
B) (1, 7)
C) (4, - 3)
D) (2, 3)
Correct Answer: B
Solution :
\[y(x)=1+\sqrt{4x-3},x>\frac{3}{4}\] Let P\[(\alpha ,\,1+\left( \sqrt{4\alpha -3} \right)\] be the point. at which \[\frac{dy}{dx}ATP=\frac{2}{3}\] \[\Rightarrow \frac{2}{\sqrt{4\alpha -3}}=\frac{2}{3}\] \[\Rightarrow 4\alpha -3=9\] \[\Rightarrow \alpha =3\] Hence P(3,4) slope of normal at P(3,4) is \[=-\frac{3}{2}\] equation of normal \[Y-4=-\frac{3}{2}(X-3)\] \[2y-8=3x+9\] \[3x+2y=17\] clearly it is passes through (1,7)You need to login to perform this action.
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