Question

equation of tangent in slope form of parabola x^2=4ay

Answer 1

dy/dx =2x/4a=x/2a. slope of tangent to given curve at point (x1,y1) =x1/2a. Equation of tangent to given curve at given point (x1,,y1). that is y-y1 =x1/2a(x-x1). in this question point is not given .

Answer 2

Equation of tangent in slope form of parabola $\frac{x^{2} }{4ay}$ is

    $y =\frac{x1*x}{2a} +\frac{x1^{2}}{2a} - y1$

find the derivative of y with respect to x, we should simplify the expression and put x in terms of y:    y = $\frac{x^{2} }{4a}$

or f(x) = $\frac{x^{2} }{4a}$

• Using differentiation, we get f’(x) = (2*x^(2–1))/4a = 2x/4a = x/2a

               So the slope of the tangent line at the point (x₁, y₁) would be                 $\frac{x1}{2a}$

• So the full equation of this tangent in standard form and solve for the y - intercept:

                            y = (x(x₁))/2a + b, where b is the y intercept

• Now let’s substitute the point (x₁, y₁) into the equation, since it the tangent line has to pass through (x₁, y₁)

                        $y1 = \frac{x1^2}{2a} + b$

                        $b = - \frac{x1^2}{2a} + y1$

• So the final equation for the tangent line to a point (x₁, y₁) is:       $y =\frac{x1*x}{2a} - \frac{x1^{2}}{2a} + y1$