Perpendicular to Tangency

Calculus Level pending

A line passing through $(-3,\ a)$ is perpendicular to a tangent line of the curve $y = x^2 - 3x + 4$ . If the point of tangency is $(3,\ 4)$ , what is the value of $a$ ?

5

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6

7

1 solution

Tom Engelsman
Nov 7, 2020

If we compute the tangent and normal lines to $y=x^2-3x+4$ at the point $(3,4)$ , we obtain:

$\frac{dy}{dx}|_{x=3} = 2(3)-3 = 3$ ,

or $y-4=3(x-3) \Rightarrow y=3x-5$ (Tangent), $y-4=-\frac{1}{3}(x-3) \Rightarrow y=-\frac{1}{3}x+5$ (Normal).

If $P(-3,a)$ is a point on the Normal Line, then $a = -\frac{1}{3}(-3)+5 = \boxed{6}.$

Perpendicular to Tangency

1 solution

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