Question

Find an equation of the line tangent to the curve y = 3x
2 − 1 and
perpendicular to the line x − 3y = 4.

Answer 1

First we have to use the definition for the curve y₁ to find m₁:

$y_1=3x^2-1 \\ y'_1=\cfrac{dy_1}{dx}=6x=m_1$

$x-3y=4 \\ 1-3y'_2=0 \implies y'_2=m_2=1/3 \\ \text{For perpendicular lines we have} \\ m_1m_2=-1 \\ \implies m_1=-1/m_2=-1/(1/3)=-3$

After we find m1 = -3, we can use the definition for y'1 to find the point (x*,y*) to find the equation of the tangent line:

$-3=m_1=6x \implies x^*=-1/2; y^*_1=3(1/2)^2-1=-1/4$

At last, we substitute to find the equation of the tangent line:

$y-y_1^*=m_1(x-x^*) \\ y-(-1/4)=(-3)(x-(-1/2)) \\ y+(1/4)=-3x-(3/2) \\ y=-3x-\cfrac{7}{4} \implies 4y+12x+7=0$

In conclusion, 4y + 12x + 7 = 0 is the equation of the tangent line to the function y = 3x² - 1 at point P ( -1/2, -1/4 ).

Reference:

• Thomas, G. B., & Finney, R. L. (1961). Calculus. Addison-Wesley Publishing Company.