Question

1. Let y = f(x) be any curve and P be any point on it then the slope of the curve at P is $\frac{dy}{dx}=m=tan\theta$
2. Equation of the tangent at P is $y-y_1=m(x-x_1)$
3. Equation of the normal at P is $y-y_1=\frac{-1}{m}(x-x_1)$
On the basis of these 3 points answer the following questions:

At what points on the curve y=$\frac{2}{3}{x}^3+\frac{1}{2}{x}^2$, tangent makes equal angle with axes

• A. $(\frac{1}{2},\frac{5}{24})$ and $(-1,-\frac{1}{6})$
• B. $(\frac{1}{2},\frac{4}{9})$ and $(-1,0)$
• C. $(\frac{1}{3},\frac{1}{7})$ and $(-3,-\frac{1}{2})$
• D. $(\frac{1}{3},\frac{4}{27})$ and $(-1,-\frac{1}{3})$

Answer 1

The correct option is A $(\frac{1}{2},\frac{5}{24})$ and $(-1,-\frac{1}{6})$

$\frac{dy}{dx}=\pm 1$
$\therefore 2x^2+x-1=0$
$x=-1,\frac{1}{2}$
$\therefore$ The points are
$(-1,-\frac{1}{6}),(\frac{1}{2},\frac{5}{24})$
$y=a{x}^2+6x+b...\left(1\right)$
In equation (1) by putting x = 0, y = 2 We get $\Rightarrow b=2$
Equation (2) ${\left|\frac{dy}{dx}\right|}_{atx=\frac{3}{2}}=0\Rightarrow 3a+6=0\Rightarrow a=-2$.