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

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(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

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must be a right triangle

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(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) :

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