Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangent to the parabola $y^2 = 6x$, parallel to $3x - 2y + 5 = 0$.

Equation of parabola is $\;$ $y^2 = 6x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 6 \implies a = \dfrac{3}{2}$

Equation of given line is $\;$ $3x - 2y + 5 = 0$

i.e. $\;$ $y = \dfrac{3}{2}x + \dfrac{5}{2}$

$\therefore$ $\;$ Slope of the given line $= m = \dfrac{3}{2}$

$\because$ $\;$ the tangent to the parabola is is parallel to the given line, slope of tangent to the parabola $= m = \dfrac{3}{2}$

The equation of tangent (with slope m) to the parabola $y^2 = 4ax$ is $\;$ $y = mx + \dfrac{a}{m}$

$\therefore$ $\;$ the required equation of tangent is

$y = \dfrac{3}{2} x + \dfrac{3 / 2}{3 / 2}$

i.e. $\;$ $y = \dfrac{3}{2}x + 1$

i.e. $\;$ $3x - 2y + 2 = 0$