Question

Let $A$ be the point where the curve $5{\alpha }^2{x}^3+10\alpha x^2+x+2y-4=0(\alpha \in R,\alpha \ne 0)$ meets the $y-$axis, then the equation of the tangent to the curve at the point where normal at $A$ meets the curve again is :

• A. $x=1$
• B. $y=2x+2$
• C. $x+2y-4=0$
• D. None of these

Answer 1

The correct option is B $y=2x+2$

$5{\alpha }^2{x}^3+10\alpha x^2+x+2y-4=0$
The curve meets $y-$axis
$\Rightarrow x=0$
So, $0+0+0+2y-4=0$
$\Rightarrow y=2$
$\therefore A(0,2)$
Now, $15{\alpha }^2{x}^2+20\alpha x+1+2\frac{dy}{dx}=0$
$\Rightarrow 0+0+1+2\frac{dy}{dx}=0$ (at point $A$)
$\Rightarrow \frac{dy}{dx}=-\frac{1}{2}$
$\therefore$ slope of normal at $A=2$
Let normal at $A$ meet the curve again at point $B(a,b)$
$\Rightarrow m_{AB}=\frac{b-2}{a-0}=2$
$\Rightarrow b=2a+2$
$\Rightarrow 2b-4=4a$
Also $(a,b)$ lies on the curve
So, $5{\alpha }^2{a}^3+10\alpha a^2+a+4a=0$
($\because 2b-4=4a$)
$\Rightarrow a(5{\alpha }^2{a}^2+10\alpha a+5)=0$
$\Rightarrow a(\alpha a+1{)}^{2}5=0$
$\Rightarrow a=-\frac{1}{\alpha }$
$\{a\ne 0\because \text{For}a=0,b=2\text{which denotes point A again}\}$
$\therefore$ Coordinates of point $B$ is $(\frac{-1}{\alpha },2-\frac{2}{\alpha })$
Now, $15{\alpha }^2{x}^2+20\alpha x+1+2\frac{dy}{dx}=0$
$\Rightarrow 15-20+1+2\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=2$(at $B$)
So, equation of tangent is
$(y-2+\frac{2}{a})=2(x+\frac{1}{a})$
$\Rightarrow y=2x+2$