Question

If the normal to the curve $y=f\left(x\right)$ at the point $(3,4)$ makes an angle $3\pi /4$ with the positive x-axis, then $f^{\prime }\left(3\right)=$

• A. $-1$
• B. $0$
• C. $1$
• D. $\sqrt{3}$

Answer 1

The correct option is C $1$

Tangent being perpendicular to given line of slope 2, will have its slope as -$\frac{1}{2}$.

Slope of tangent=$-\frac{fx}{fy}=-\frac{6x+1}{2(y+1)}=-\frac{1}{2}\therefore y=6x$.

Sloping with the given curve, we have $3x^2+36x^2+x+12x=0$

or $13x(3x+1)=0\therefore x=0,-1/3\therefore y=0,-2$

Hence the two points are $(0,0),(-\frac{1}{3},-2)$ $\therefore y=-\frac{1}{2}x$ and $y+2=-\frac{1}{2}(x+\frac{1}{3})$ or $2y+x=0$ and $2y+x+\frac{13}{3}=0$

Ans: D