If A(at^2, 2at),B(a/t^2, -2a/t),and C(a, 0),t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given points are $A(at^2, 2at)$,$B(a/t^2, -2a/t)$,and $C(a, 0)$.$CA = \sqrt{(at^2 - a)^2 + (2at - 0)^2} = \sqrt{a^2(t^2 - 1)^2 + 4a^2t^2} = a\sqrt

Vedclass · Accepted Answer

$H.M.$ of $CA$ and $CB$

Vedclass · Answer

(C) Given points are $A(at^2, 2at)$,$B(a/t^2, -2a/t)$,and $C(a, 0)$.$CA = \sqrt{(at^2 - a)^2 + (2at - 0)^2} = \sqrt{a^2(t^2 - 1)^2 + 4a^2t^2} = a\sqrt{t^4 - 2t^2 + 1 + 4t^2} = a\sqrt{(t^2 + 1)^2} = a(t^2 + 1)$.$CB = \sqrt{(\frac{a}{t^2} - a)^2 + (-\frac{2a}{t} - 0)^2} = \sqrt{a^2(\frac{1 - t^2}{t^2})^2 + \frac{4a^2}{t^2}} = a\sqrt{\frac{1 - 2t^2 + t^4 + 4t^2}{t^4}} = a\sqrt{\frac{(1 + t^2)^2}{t^4}} = a\frac{1 + t^2}{t^2}$.$H.M.$ of $CA$ and $CB$ is given by $\frac{2(CA)(CB)}{CA + CB}$.$H.M. = \frac{2 \cdot a(t^2 + 1) \cdot a(\frac{1 + t^2}{t^2})}{a(t^2 + 1) + a(\frac{1 + t^2}{t^2})} = \frac{2a^2(t^2 + 1)^2 / t^2}{a(t^2 + 1)(1 + 1/t^2)} = \frac{2a(t^2 + 1)^2 / t^2}{(t^2 + 1)(\frac{t^2 + 1}{t^2})} = \frac{2a(t^2 + 1)^2 / t^2}{(t^2 + 1)^2 / t^2} = 2a$.Thus,$2a$ is the $H.M.$ of $CA$ and $CB$.

If $A(at^2, 2at)$,$B(a/t^2, -2a/t)$,and $C(a, 0)$,then $2a$ is equal to

Similar Questions

The centroid of the triangle formed by joining the feet of the normals drawn from any point to the parabola $y^2 = 4ax$ lies on

If $x-2=t^2$ and $y=2t$ are the parametric equations of the parabola $y^2=a(x-b)$,then the value of $a+b$ equals

If the normal at the point $(bt_1^2, 2bt_1)$ on the parabola $y^2 = 4bx$ meets the parabola again at the point $(bt_2^2, 2bt_2)$,then:

If the vertex of a parabola is $(2, -1)$ and the equation of its directrix is $4x - 3y = 21$,then the length of its latus rectum is

The $Y$-intercept of the common tangent to the parabola $y^2 = 32x$ and $x^2 = 108y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module