Question

Which of the following statements is false?

• A. The length of sub-tangent to the curve $x^2{y}^2=16a^4$ at the point $(-2a,2a)$ is $2a$
• B. $x+y=3$ is a normal to the curve $x^2=4y$
• C. Curves $y=-4x^2$ and $y=e^{-x/2}$ are orthogonal
• D. If a $\in (-1,0),$ then tangent at each point of the curve $y=\frac{2}{3}{x}^3-2a{x}^2+2x+5$ makes an acute angle with the positive direction of x-axis

Answer 1

Answer: (C)

$\left(a\right)x^2{y}^2=16a^4$
$L_{ST}=\left|\frac{y}{x_T}\right|\Rightarrow xy=4a^2$
$y+x{y}^{\prime }=0$
$y^{\prime }=\frac{-y}{x};L_{ST}=\left|\frac{y}{y/x}\right|=x\Rightarrow L_{ST}=2a$
$\therefore \left(a\right)$ is true.
$\left(b\right)\frac{dy}{dx}=\frac{x}{2}=1i.e.x=2$
$\therefore \left(2,1\right)$ is the point on the curve $x^2=4y$ at which the normal is
$y-1=-1\left(x-2\right)i.e.x+y=3\therefore \left(b\right)$ is true
$\left(c\right)y=-4x^2,y=e^{-x/2}$
The curves are non-intersecting
$\therefore$ curves are not orthogonal i.e. $\left(c\right)$ is false.
$\left(d\right)y=\frac{2}{3}{x}^3-2a{x}^2+2x+5$
$\frac{dy}{dx}=2x^2-4ax+2\left(x^2-2ax+1\right)$
$=2{\left(x-a\right)}^2+2-2a^2>0$
$\left[\because a\in \left(-1,0\right)\Rightarrow 0<a^2<1\right]$
$\therefore \left(d\right)$ is true