Question

Find equations of both lines through the point (2, −3) that are tangent to the parabola y = x^2 + x.

Linear equations and graphs
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asked 2021-05-23
Find equations of both lines through the point (2, −3) that are tangent to the parabola \(y = x^2 + x\).

Answers (1)

2021-05-24

Note that all points on the parabola \(y = x^2 + x\) are of the form \((x, x^2 + x)\)
Assume that the tangent at \((a, a^2 + a)\) passes through (2, -3)
We know that the slope of tangent at any point is the derivative at that point.
Differentiate \(y = x^2 + x, To\ get \frac{dy}{dx} = 2x + 1\)
Therefore, the slope of tangent at \((a, a^2 + a) is 2a + 1 \rightarrow (1)\)
Since the tangent passes through the points \((a, a^2 + a)\) and (2. -3). We can write the slope of the tangent as
\(\frac{(a^2 + a)-(-3)}{a - 2} = \frac{a^2 + a + 3}{a - 2} \rightarrow (2)\)
Using (1) and (2), we can write \(\frac{a^2 + a + 3}{a - 2} = 2a + 1\)
\(a^2 + a + 3 = (2a + 1)(a - 2)\)
\(a^2 + a + 3 = 2a^2 + a - 4a - 2\)
\(a^2 - 4a - 5 = 0\)
\((a + 1)(a - 5) = 0 \Rightarrow a = -1\ and\ a = 5\)
Therefore, there are two tangents that pass through (2, -3)
Finding the equation of the tangent corresponding to \(a = -1\)
Note that the slope of the tangent is \(2a + 1 = 2(-1) + 1 = -1\)
Since the tangent. passes through (2, -3).
Therefore, the equation of the tangent is
\(\frac{y - (-3)}{x - 2} = -1\)
\(y + 3 = -(x - 2)\)
\(x + y + 1 = 0\)
Finding the equation of the tangent corresponding to \(a = 5\)
Note that the slope of the tangent is \(2a + 1 = 2(5) + 1 = 11\)
Since the tangent. passes through (2, -3).
Therefore, the equation of the tangent is
\(\frac{y - (-3)}{x - 2} = 11\)
\(y + 3 = 11(x - 2)\) \(y = 11x - 25\) image

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