Find equations of both lines through the point (2, −3) that are tangent to the parabola y=x2+x.

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Clara Reese · Accepted Answer

Note that all points on the parabola y=x2+x are of the form (x,x2+x) Assume that the tangent at (a,a2+a) passes through (2, -3) We know that the slope of tangent at any point is the derivative at that point. Differentiate y=x2+x,To getdydx=2x+1 Therefore, the slope of tangent at (a,a2+a)is2a+1→(1) Since the tangent passes through the points (a,a2+a) and (2. -3). We can write the slope of the tangent as (a2+a)−(−3)a−2=a2+a+3a−2→(2) Using (1) and (2), we can write a2+a+3a−2=2a+1 a2+a+3=(2a+1)(a−2) a2+a+3=2a2+a−4a−2 a2−4a−5=0 (a+1)(a−5)=0⇒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 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 y−(−3)x−2=11 y+3=11(x−2) y=11x−25

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

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