Question

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
I. x2 + xy - y2 = 1, (2,3) (10)
II x2 + y2 = 25, (3,-4)
III. x3y2 = 9, (-1,3)
IV. y2 – 2x – 4y - 1 = 0, (-2, 1)

Answer 1

For each curve, plug in the given point $(x,y)$ and check if the equality holds. For example:

(I) (2, 3) does lie on $x^2+xy-y^2=1$ since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative $\frac{\mathrm dy}{\mathrm dx}$, and evaluate it for the given point $(x,y)$. This is the slope of the tangent line at the point. For example:

(I) The derivative is

$x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}$

so the slope of the tangent at (2, 3) is

$\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74$

and its equation is then

$y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12$

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, $-\frac1{\frac{\mathrm dy}{\mathrm dx}}$. For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

$y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7$