Find the coordinates of the points where the tangent of the curve $y=3x^2-2x-4$ are perpendicular to the line $x+10y-7=0$

Question

Find the coordinates of the points where the tangent of the curve $y=3x^2-2x-4$ are perpendicular to the line $x+10y-7=0$

So, $$\frac{dy}{dx}=6x-2$$ ------- (1)

Differentiate second one respect to $x$

$$1+10 \frac{dy}{dx}-7=0$$ $$\frac{dy}{dx}=-\frac{1}{10}$$ ----------(2)

If tangent of 1 is perpendicular to 2 than.

$$-\frac{1}{10} (6x-2)=-1$$

That's what book wrote. But, when I wrote above equation. I found something just like this

$$-\frac{1}{10}=6x-2$$

I was trying to get the same value. But, I didn't get it. Why? I think I did a typo mistake. But, I can't find it.

Answer 1

All lines perpendicular to the $x+10y-7=0$, whose angular coefficient $m'=-\frac{a}{b} =-\frac{1}{10}$ have equation

$y=m x+q$,

where

$m=-\frac{1}{m'}=10$, ( for the condition of perpendicularity)

and $q$ is the segment intercepted on the $y$ axis.

We must remember that the angular coefficient of the tangent to a curve is equal to the value of the first derivative of the equation of the curve, calculated at the point of tangency. Therefore,

$\frac{dy}{dx}=6x-2$,

or

$10=6x-2$

This condition allows us to calculate $x=2$.

Line $y=10x+q$ and curve $y=3x^{2}-2x-4$ must have at least one point in common, so equalizing the ordinates $y$ we get: $10x+q=3x^{2}-2x-4$ This is an equation of second degree, and for the tangency condition the two points must coincide, i.e. be only one. This results in imposing the discriminant of the equation $\Delta=b^{2}-4ac=0$, (a,b,c) being the coefficients of the equation of second degree $ax^{2}+bx+c=0$. We arrive at:

$3x^{2}-12x-q-4=0$

$\Delta=6^{2}+3q+12=0$,

$q=-16$.

The equation of the line is $y=10x-16$

and the coordinates of the tangency point are (2,4).

Answer 2

$$x+10y-7=0,y=-\frac{1}{10}x+\frac{7}{10}$$

Since lines are perpendicular $m_1\times m_2 =-1$, So slope of line is $m=10$

$f'(x)=6x-2=10, x=2$ and $f(2)=4$

Equation of line is

$y-4=10(x-2), y=10x-16$