If the tangent to y^{2}=4ax at the point (at^ | vedclass

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

Question

(A) The equation of the tangent to the parabola $y^{2}=4ax$ at the point $(at^{2}, 2at)$ is given by $ty = x + at^{2}$, which can be rewritten as $x -

Vedclass · Accepted Answer

$t=-\operatorname{cosec} 	heta$

Vedclass · Answer

(A) The equation of the tangent to the parabola $y^{2}=4ax$ at the point $(at^{2}, 2at)$ is given by $ty = x + at^{2}$, which can be rewritten as $x - ty + at^{2} = 0$. The equation of the normal to the hyperbola $x^{2}-y^{2}=a^{2}$ at the point $(a \sec 	heta, a 	an 	heta)$ is given by $\frac{ax}{\sec 	heta} + \frac{ay}{	an 	heta} = a^{2} + a^{2}$, which simplifies to $x \cos 	heta + y \cot 	heta = 2a$. Comparing the two equations $x - ty + at^{2} = 0$ and $x \cos 	heta + y \cot 	heta - 2a = 0$, we have the ratios of coefficients: $\frac{1}{\cos 	heta} = \frac{-t}{\cot 	heta} = \frac{at^{2}}{-2a}$. From $\frac{1}{\cos 	heta} = \frac{-t}{\cot 	heta}$, we get $t = -\frac{\cot 	heta}{\cos 	heta} = -\operatorname{cosec} 	heta$. From $\frac{1}{\cos 	heta} = \frac{at^{2}}{-2a}$, we get $t^{2} = -2 \sec 	heta$.

If the tangent to $y^{2}=4ax$ at the point $(at^{2}, 2at)$ where $|t|>1$ is a normal to $x^{2}-y^{2}=a^{2}$ at the point $(a \sec \theta, a \tan \theta)$, then

Similar Questions

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. The equation of the common tangent with a positive slope to the circle and the hyperbola is:

Let $P$ be a moving point on the circle $x^2 + y^2 - 6x - 8y + 21 = 0$. Then, the maximum distance of $P$ from the vertex of the parabola $x^2 + 6x + y + 13 = 0$ is equal to:

The normal at $\left( 2, \frac{3}{2} \right)$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ touches a parabola,whose equation is

The number of points of intersection of $|z - (4 + 3i)| = 2$ and $|z| + |z - 4| = 6$,$z \in \mathbb{C}$ is

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=8ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $e^4$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module