Question

Find the coordinates of the point at which the normal to the curve y = x ^ 2 at x = 1 meets the curve again

Answer 1

Answer:

$(-\frac{3}{2},\frac{9}{4})$

Step-by-step explanation:

A normal line is a line perpendicular to the tangent line, so we must take the derivative of the function to find the slope of the tangent line, and then take the opposite reciprocal of this slope, to find the slope of the normal line. The equation of a normal line will be in the form of $y=mx+b$:

$f(x)=x^2\\\\f'(x)=2x\\\\f'(1)=2(1)\\\\f'(1)=2$

Hence, the opposite reciprocal is $-\frac{1}{2}$, which is our slope for the normal line. Now, we find the y-intercept by plugging in the given coordinate point $(1,1)$ since $f(1)=1^2=1$:

$y=-\frac{1}{2}x+b\\\\1=-\frac{1}{2}(1)+b\\ \\1=-\frac{1}{2}+b\\ \\b=\frac{3}{2}$

Thus, the equation of the normal line is $y=-\frac{1}{2}x+\frac{3}{2}$. Using this information, we can now find the coordinates where the normal of the curve at x=1 meets the curve again:

$x^2=-\frac{1}{2}x+\frac{3}{2}\\ \\ x^2+\frac{1}{2}x-\frac{3}{2}=0\\ \\ 2x^2+x-3=0\\\\(x-1)(2x+3)=0\\\\x=1,\: x=-\frac{3}{2}$

Since we only care about $x=-\frac{3}{2}$ since $x=1$ was already accounted for, the y-coordinate would be:

$y=x^2\\\\y=(-\frac{3}{2})^2\\\\y=\frac{9}{4}$

Therefore, the coordinates are $(-\frac{3}{2},\frac{9}{4})$.

In the graph attached below, you can see the tangent line $y=2x-1$ and how it is perpendicular to $y=\frac{1}{2}x-\frac{3}{2}$.