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

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Theirl1972 · Accepted Answer

Given:  The given function is =x2+x  Find the derivative:  =y=x2+x  =dydx=2x+1  Find the equation of the tangent line:  Since y=x2+x. Considerthe point on the porabola (a,a2+a)  Thus the slope of the line at (a,a2+a) is  m=dydx=2a+1  By using the slope intercept form =y=mx+c⇒y=(2a+1)x=c  If the line is a tangent to a given parabola,then it passes through (a,a2+a)  =a2+a=(2a+1)+c  =a2+a=2a2+1+c  =c=−a2  Thus the equation of the langent line is =y=(2a+1)x−a2.  The required tangent passes through the point (2,−3)  Substitute =x=2 and =y=−3 in above equation  =−3=(2a+1)2−a2  =−3=4a+2−a2  =a2−4a−5=0  =(a+1)(a−5)=0  =a=−1 or =a=5  Substitute =a=−1 in =y=(2a+1)x−a2  =y=(2(−1)+1)x−(−1)2)  =y=−x−1  Substitute =a=5 in y=(2a+1)x−a2  =y=(2(5)+1)x−(5)2  =y=11x−25

Find equations of both lines through the point (2, -3)

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