Find the equation of the tangent to the parabola y2 = 8x which is para

1.	Find the equation of the tangent to the parabola y2 = 8x which is parallel to the line 2x + 2y + 5 = 0. Find its point of contact.
Answer» Given the equation of the parabola is y² = 8x. Comparing this equation with y² = 4ax, we get 4a = 8 a = 2 Slope of the line 2x + 2y + 5 = 0 is -1 Since the tangent is parallel to the given line, slope of the tangent line is m = -1 Equation of tangent to the parabola y² = 4ax having slope m is y = mx + a/m Equation of the tangent is y = -x + 2/-1 x + y + 2 = 0 Point of contact =(\(\frac{a}{m^2}, \frac{2a}{m}\)) = \(\frac {2}{(-1)^2}, \frac{2(2)}{-1}\) = (2, -4)

Find the equation of the tangent to the parabola y2 = 8x which is parallel to the line 2x + 2y + 5 = 0. Find its point of contact.

Answer»

Given the equation of the parabola is y² = 8x.

Comparing this equation with y² = 4ax, we get

4a = 8

a = 2

Slope of the line 2x + 2y + 5 = 0 is -1

Since the tangent is parallel to the given line, slope of the tangent line is m = -1

Equation of tangent to the parabola y² = 4ax having slope m is

y = mx + a/m

Equation of the tangent is

y = -x + 2/-1

x + y + 2 = 0

Point of contact =(\(\frac{a}{m^2}, \frac{2a}{m}\))

= \(\frac {2}{(-1)^2}, \frac{2(2)}{-1}\)

= (2, -4)