1) Find the point P on the parabola y^2 = 4ax such that area parabola, the X-axis and the tangent - Maths - Application of Integrals

Question

1) Find the point P on the parabola y^2 = 4ax such that area
parabola, the X-axis and the tangent at P is equal to that of
bounded by the parabola, the X-axis and the normal at P
2) Find the area of the region bounded by C: y = tanx , tangent
drawn at C at x = pie/4 and the x- axis

Himanshu · Accepted Answer

$y = \tan x \frac{d y}{d x} = \sec^{2} x m = \sec^{2} (\frac{π}{4}) = 2 y - 1 = 2 (x - \frac{π}{4}) \Rightarrow y - 2 x = 1 - \frac{π}{2} A r e a b o u n d e d = | \int_{0}^{\frac{π}{4}} \tan x ⅆ x - \int_{\frac{π}{4} - \frac{1}{2}}^{\frac{π}{4}} (2 x + 1 - \frac{π}{2}) ⅆ x | A = | \ln \sec \frac{π}{4} - {(x^{2} + x - \frac{π x}{2})}_{\frac{π}{4} - \frac{1}{2}}^{\frac{π}{4}} | A = \frac{1}{2} \ln (2) - \frac{1}{4}$

$P (a t^{2}, 2 a t) A_{1} = \int_{- a t^{2}}^{a t^{2}} (\frac{x}{t} + a t) ⅆ t - \int_{0}^{a t^{2}} 2 \sqrt{a x} ⅆ x A_{1} = 2 a^{2} t^{3} - \frac{4}{3} a^{2} t^{3} = \frac{2}{3} a^{2} t^{3} A_{2} = \int_{0}^{a t^{2}} 2 \sqrt{a x} ⅆ x + \int_{a t^{2}}^{2 a + a t^{2}} (- t x + 2 a t + a t^{3}) ⅆ x A_{2} = \frac{4}{3} a^{2} t^{3} + 2 a^{2} t A_{1} = A_{2} \Rightarrow - \frac{2}{3} a^{2} t^{3} = 2 a^{2} t \Rightarrow t^{2} = - \frac{3}{2} T h i s d o e s n o t g i v e a n y r e a l v a l u e o f t .$

1) Find the point P on the parabola y^2 = 4ax such that area parabola, the X-axis and the tangent at P is equal to that of bounded by the parabola, the X-axis and the normal at P. 2) Find the area of the region bounded by C: y = tanx , tangent drawn at C at x = pie/4 and the x- axis.

1) Find the point P on the parabola y^2 = 4ax such that area parabola, the X-axis and the tangent at P is equal to that of bounded by the parabola, the X-axis and the normal at P.

2) Find the area of the region bounded by C: y = tanx , tangent drawn at C at x = pie/4 and the x- axis.