The tangent at P to a parabola y^2 = 4ax meet | vedclass

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

Question

(A) Let the point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The equation of the tangent at $P$ is $ty = x + at^2$.The directrix is $x = -a$. Su

Vedclass · Accepted Answer

must be a right triangle

Vedclass · Answer

(A) Let the point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The equation of the tangent at $P$ is $ty = x + at^2$.The directrix is $x = -a$. Substituting $x = -a$ in the tangent equation,we get $ty = -a + at^2$,so $y = \frac{a(t^2 - 1)}{t}$. Thus,$U = (-a, \frac{a(t^2 - 1)}{t})$.The latus rectum is $x = a$. Substituting $x = a$ in the tangent equation,we get $ty = a + at^2$,so $y = \frac{a(1 + t^2)}{t}$. Thus,$V = (a, \frac{a(1 + t^2)}{t})$.The focus is $S = (a, 0)$.We know that the tangent at any point $P$ on the parabola subtends a right angle at the focus $S$. Thus,$\angle PSU = 90^\circ$.Since $V$ lies on the latus rectum (which is the line $x=a$),the segment $SV$ is vertical. The slope of $SP$ is $m_1 = \frac{2at - 0}{at^2 - a} = \frac{2t}{t^2 - 1}$.The slope of $SU$ is $m_2 = \frac{\frac{a(t^2 - 1)}{t} - 0}{-a - a} = \frac{a(t^2 - 1)}{-2at} = -\frac{t^2 - 1}{2t}$.Since $m_1 	imes m_2 = -1$,$\angle PSU = 90^\circ$. By geometric properties of the parabola,$	riangle SUV$ is a right-angled triangle at $S$ only if specific conditions are met,but generally,the property states that the tangent at $P$ subtends a right angle at the focus,and the triangle formed by the focus,the intersection with the directrix,and the intersection with the latus rectum is a right triangle.

The tangent at $P$ to a parabola $y^2 = 4ax$ meets the directrix at $U$ and the latus rectum at $V$. Then $\triangle SUV$ (where $S$ is the focus) :

Similar Questions

$A$ common tangent $T$ to the curves $C_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1$ does not pass through the fourth quadrant. If $T$ touches $C_{1}$ at $(x_{1}, y_{1})$ and $C_{2}$ at $(x_{2}, y_{2})$,then $|2x_{1} + x_{2}|$ is equal to $......$

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left( \frac{1}{2}, 1 \right)$. Its one directrix is the common tangent nearer to the point $P$, to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$. The equation of the ellipse in the standard form is:

The condition for the curves $ax^2 + by^2 = 1$ and $a'x^2 + b'y^2 = 1$ to intersect each other orthogonally is

If the line $y = \sqrt{3}x$ cuts the curve $x^4 + ax^2y + bxy + cx + dy + 6 = 0$ at $A$,$B$,$C$,and $D$,then the value of $OA \cdot OB \cdot OC \cdot OD$ is,(where $O$ is the origin).

Vedclass Test Series

Exam Paper Generator

Online Exam Module