Answer to Question #267697 in Calculus for kong

Question

Find the tangent to the parabola y² = 6x − 3 perpendicular to the line x + 3y = 7

Answer 1

Tangent line to the parabola y²=6x-3,

$2yy'=6\\ y'=\frac{3}{y}\\$

And slope of the straight line x+3y=7 is $\frac{-1}{3}$ .

$y=\frac{-1}{3}x+\frac{7}{3}$

Since, tangent to the parabola is perpendicular to the straight line.

Therefore, product of slope of two curves is -1.

$m_s.m_p=-1\\ \frac{3}{y}.\frac{-1}{3}=-1\\ y=1$

Now, by putting y=1 in the parabola gives the tangent point.

$1^2=6x-3\\ x=\frac{2}{3}$

Thus, tangent point is $(\frac{2}{3},1)$ .

And, the equation of the tangent line is the line passing $(\frac{2}{3},1)$ and having slope 3.

i.e.,

$y=mx+b\\ 1=3.\frac{2}{3}+b\\ b=-1$

Therefore, equation of tangent line is y=3x-1.