If a normal chord at a point t (neq 0) on the | vedclass

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

Question

(D) The equation of the given parabola is $y^2 = 9x$. Here,$4a = 9$,so $a = \frac{9}{4}$.The equation of the normal chord at point $t$ is $y = -tx + 2

Vedclass · Accepted Answer

$\pm \sqrt{2}$

Vedclass · Answer

(D) The equation of the given parabola is $y^2 = 9x$. Here,$4a = 9$,so $a = \frac{9}{4}$.The equation of the normal chord at point $t$ is $y = -tx + 2at + at^3$.Substituting $a = \frac{9}{4}$,we get $y = -tx + \frac{9}{2}t + \frac{9}{4}t^3$,which can be written as $tx + y = \frac{9}{2}t + \frac{9}{4}t^3$.Since the chord subtends a right angle at the vertex $V(0,0)$,we homogenize the parabola $y^2 = 9x$ using the line equation:$y^2 = 9x \left( \frac{tx + y}{\frac{9}{2}t + \frac{9}{4}t^3} ight)$$y^2 = 9x \left( \frac{tx + y}{\frac{9}{4}t(2 + t^2)} ight) = \frac{4x(tx + y)}{t(2 + t^2)}$$t(2 + t^2)y^2 = 4tx^2 + 4xy$$4tx^2 + 4xy - t(2 + t^2)y^2 = 0$For a right angle at the origin,the sum of coefficients of $x^2$ and $y^2$ must be zero:$4t - t(2 + t^2) = 0$Since $t 
eq 0$,we divide by $t$:$4 - (2 + t^2) = 0$$4 - 2 - t^2 = 0$$t^2 = 2 \Rightarrow t = \pm \sqrt{2}$.Thus,option $D$ is correct.

If a normal chord at a point $t (\neq 0)$ on the parabola $y^2 = 9x$ subtends a right angle at its vertex,then $t =$

Similar Questions

If the chord joining the points $P_{1}(x_{1}, y_{1})$ and $P_{2}(x_{2}, y_{2})$ on the parabola $y^{2} = 12x$ subtends a right angle at the vertex of the parabola, then $x_{1}x_{2} - y_{1}y_{2}$ is equal to

If the length of the latus rectum of a parabola,whose focus is $(a, a)$ and the tangent at its vertex is $x+y=a$,is $16$,then $|a|$ is equal to.

If $P$ is a point on the parabola $y^{2}=4ax$ with focus $F$. Let $Q$ denote the foot of the perpendicular from $P$ onto the directrix. Then, $\frac{\tan \angle PQF}{\tan \angle PFQ}$ is

If $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$,then the slope of the tangent drawn at $(1, k)$ on the given parabola is

The angle of intersection between the curves $x^2 = 8y$ and $y^2 = 8x$ at the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module