Let C be a curve given by $y=1+\sqrt{4 x-3}$, $x>\frac{3}{4}$.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
Answer & explanation
Correct answer: option 2
Let $P(\alpha, 1+\sqrt{4 \alpha-3})$ be a point on C, where $\alpha>\frac{3}{4}$ such that tangent at P has slope $\frac{2}{3}$.
Now, $y=1+\sqrt{4 x-3}$
$\Rightarrow \frac{d y}{d x}=\frac{2}{\sqrt{4 x-3}} \Rightarrow\left(\frac{d y}{d x}\right)_P=\frac{2}{\sqrt{4 \alpha-3}}$
It is given that $\left(\frac{d y}{d x}\right)_P=\frac{2}{3}$
$\Rightarrow \frac{2}{\sqrt{4 \alpha-3}}=\frac{2}{3} \Rightarrow 4 \alpha-3=9 \Rightarrow \alpha=3$
So, the coordinates of P are (3, 4)
The equation of the normal at P(3, 4) is
$y-4=-\frac{3}{2}(x-3)$ or, $3 x+2 y-17=0$
Clearly, it basses through (1, 7).