Question

Find the coordinates of the points on the curve $y=x^2+3x+4$, the tangents at which pass through the origin.

Answer 1

Step 1: Differentiate the given curve to obtain slope of tangents

$y=x^2+3x+4$

Differentiating with respect to $x$ we get,

$\begin{array}{rcl}{\frac{dy}{dx}}_{\left|x,y\right|}& =& \frac{d}{dx}(x^2+3x+4)\\ m& =& 2x+3\end{array}$

Step 2: Write the equation of tangents

Let $\left(x_{1,}{y}_1\right)$ be a point on the curve

As the tangents pass through the origin, the equation of the tangent can be written as

$y=mx$

Substitute the values of the co-ordinates we get,

$y_1=m{x}_1$

At point $\left(x_{1,}{y}_1\right)$ the value of $m$ is given as,

$m={\frac{dy}{dx}}_{\left|x_{1,}{y}_1\right|}=2x_1+3$

$\Rightarrow$ $y_1=\left(2x_1+3\right)x_1$

$\Rightarrow$$y_1=\left(2{x^2}_1+3x_1\right)$ $...\left(i\right)$

Step 3: Substitute the co-ordinates in the equation of the curve

$\Rightarrow$ $y_1={x_1}^2+3x_1+4$ $...\left(ii\right)$

From $\left(i\right),\left(ii\right)$ we get,

$2{x_1}^2+3x_1={x_1}^2+3x_1+4$

$\Rightarrow$ ${x_1}^2-4=0$

$\Rightarrow$ $\left(x_1-2\right)\left(x_1+2\right)=0$

$\Rightarrow$ $x_1=2$ and $x_1=-2$

Resubstituting the value of $x_1$in $\left(i\right)$ we get,

$\Rightarrow$ $y_1=2\times 2^2+3\times 2$ and $y_1=2\times {\left(-2\right)}^2+3\times \left(-2\right)$

$\Rightarrow$ $y_1=14$ and $y_1=2$

Hence, the ordered pairs are $\left(2,14\right)$ and $\left(-2,2\right)$

Hence, the co-ordinates of the points on the curve $y=x^2+3x+4$ at which tangents pass through the origin are $\left(2,14\right)$ and $\left(-2,2\right)$