Question

The coordinates of the point on the curve $y=x^2+3x+4$ the tangent at which passes through the origin are -

• A. $(-2,2),(2,14)$
• B. $(1,-1),(3,4)$
• C. $(2,14),(2,2)$
• D. $(1,2),(14,3)$

Answer 1

Answer: (A)

$(-2,2),(2,14)$

Let the point is $P(a,b)$
Now the given curve is $y=x^2+3x+4$
Differentiating w.r.t $x$
$\frac{dy}{dx}=2x+3$
Thus slope of tangent at P is $={\left(\frac{dy}{dx}\right)}_{(a,b)}=2a+3$
Hence equation of tangent at P is given by, $(y-b)=(2a+3)(x-a)$
But given that tangent passes through origin
$\Rightarrow (0-b)=(2a+3)(0-a)\Rightarrow 2a^2+3a=b..\left(1\right)$
Also the point P lies on the given curve,
$b=a^2+3a+4...\left(2\right)$
Solving (1) and (2) we get the required point P coordinate
which are $(-2,2)\text{or}(2,14)$
Hence, option 'A' is correct.