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A tangent to the curve \[y = {x^2} + 3x\] passes through a point \[(0, - 9)\], find the points.

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Last updated date: 17th Jun 2024
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Answer
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Hint: Using the equation of the curve we will find the slope of the tangent at a certain point. Then, since the point passes through the curve and the tangent, it satisfies the equation of the curve and the tangent. Solving these two equations we can find the required points.

Complete step-by-step answer:
It is given that; the equation of the curve is \[y = {x^2} + 3x\]. The curve passes through the point \[(0, - 9)\]. We have to find the point of tangent.
Let us consider, the curve has the tangent at the point \[(h,k)\].
To find the slope of the tangent, we will differentiate the curve with respect to \[x\].
Let us differentiate the curve \[y = {x^2} + 3x\], with respect to \[x\] we get,
Then the slope of the given curve is \[\dfrac{{dy}}{{dx}} = 2x + 3\]
So, the slope of the tangent at \[(h,k)\] is \[2h + 3\]
Now to find the points we need to find the equation of the tangent line.
We know that the equation of the tangent line passes through the point \[{x_1}, {y_1}\] is \[y - {y_1} = m(x - {x_1})\] where m is the slope of the tangent.
Here it is given that the tangent line passes through \[(0, - 9)\].
Therefore, the equation the tangent line passes through the \[(0, - 9)\], so,
\[y + 9 = (2h + 3)(x - 0)\]
On simplifying the above equation we get,
\[y = 2hx + 3x - 9\]
Again, the point \[(h,k)\] passes through the curve and the tangent, by substituting the points in the curve and tangent we get,
\[k = {h^2} + 3h\]… (1)
\[k = 2{h^2} + 3h - 9\]… (2)
By equating (1) and (2) we get,
\[{h^2} + 3h = 2{h^2} + 3h - 9\]
Simplifying we get,
\[{h^2} = 9\]
Let us take square root on both sides, then we get,
\[h = \pm 3\]
Taking, \[h = 3\] we get,
\[k = {3^2} + 3 \times 3 = 18\]
Taking, \[h = - 3\] we get,
\[k = {3^2} + (3 \times - 3) = 0\]
Hence, the required points are, \[(3,18)\] $and$ \[( - 3,0).\]

Note: In geometry, the tangent line or simply tangent to a curve at a given point is the straight line that just touches the curve at a point.
It is given by the formula \[y - {y_1} = m(x - {x_1})\] where m is the slope and \[{y_1}\] $and$ \[{x_1}\] are the points through which the curve passes.
When a point passes through a curve or any line, it satisfies the equation of the curve or the line.