The point of intersection of tangents at the | vedclass

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

Question

(B) The parabola is given by $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$.The ends of the latus rectum are $L(a, 2a)$ and $L_1(a, -2a)$,

Vedclass · Accepted Answer

$(-1,0)$

Vedclass · Answer

(B) The parabola is given by $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$.The ends of the latus rectum are $L(a, 2a)$ and $L_1(a, -2a)$,which are $L(1, 2)$ and $L_1(1, -2)$.The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.For point $L(1, 2)$,the tangent equation is $y(2) = 2(1)(x + 1) \implies y = x + 1$.For point $L_1(1, -2)$,the tangent equation is $y(-2) = 2(1)(x + 1) \implies -y = x + 1$.Adding the two equations: $(x + 1) + (x + 1) = 0 \implies 2x + 2 = 0 \implies x = -1$.Substituting $x = -1$ into $y = x + 1$,we get $y = -1 + 1 = 0$.Thus,the point of intersection is $(-1, 0)$.

The point of intersection of tangents at the ends of the latus rectum of the parabola $y^2 = 4x$ is equal to

Similar Questions

If $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to:

The minimum value of $|z-1|+|z-5|$ is

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is $60^{\circ}$,then the coefficient of friction is

If $y$ is a function of $x$ and $\log(x+y) = 2xy$,then the value of $y'(0)$ is

$A$ fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self-inductance $L$ at $t = 0$. The time at which the energy is stored equally between the electric and the magnetic fields is

Vedclass Test Series

Exam Paper Generator

Online Exam Module