Question

A point on the curve $y=2x^3+13x^2+5x+9$, the tangent at which passes through the origin is

• A. $(1,15)$
• B. $(1,-15)$
• C. $(15,1)$
• D. $(-1,15)$

Answer 1

The correct option is D $(-1,15)$

Let $P(a,b)$ be a point on the given curve $y=2x^3+13x^2+5x+9$

$b=2a^3+13a^2+5a+9$ --------- ( 1 )

Now, taking derivative of $y$.

$\therefore$ $y^{\prime }=6x^2+26x+5\Rightarrow y^{\prime }atP=6a^2+26a+5$.

Equation of tangent at $P$ is

$y-b=(6a^2+26a+5)(x-a)$

This tangent passes through origin means

$b=a(6a^2+26a+5)$

$2a^3+13a^2+5a+9=6a^3+26a^2+5a$

$4a^3+13a^2-9=0$

$(a+1)(a+3)(4a-3)=0$

Taking, $a+1=0$

We get, $a=-1$

Substituting $a=-1$ in equation (1) we get,

$b=2(-1{)}^3+13(-1{)}^2+5(-1)+9$

$\therefore$ $b=15$

$\therefore$ $P(a,b)=(-1,15)$

$\therefore$ A point on the curve $y=2x^3+13x^2+5x+9$, the tangent at which passes through the origin is $(-1,15)$.