Variable chords of the parabola y^2 = 4ax sub | vedclass

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

Question

(D) Let the chord be $y = mx + c$. Since it passes through the parabola $y^2 = 4ax$,the combined equation of the lines joining the vertex $(0, 0)$ to

Vedclass · Accepted Answer

all of the above

Vedclass · Answer

(D) Let the chord be $y = mx + c$. Since it passes through the parabola $y^2 = 4ax$,the combined equation of the lines joining the vertex $(0, 0)$ to the points of intersection is $y^2 = 4ax \left(\frac{y-mx}{c}\right)$.Since the chord subtends a right angle at the vertex,the sum of the coefficients of $x^2$ and $y^2$ is $0$.$1 + \frac{4am}{c} = 0 \implies c = -4am$.Thus,the chord is $y = mx - 4am$,which always passes through the fixed point $(4a, 0)$.The locus of the feet of the perpendiculars from the vertex to the chord $y = mx - 4am$ is $x^2 + y^2 - 4ax = 0$,which is a circle.The locus of the midpoints of the chords is $y^2 = 2a(x - 4a)$,which is a parabola.Therefore,all the given statements are correct.

Variable chords of the parabola $y^2 = 4ax$ subtend a right angle at the vertex. Then:

Similar Questions

The equation of a tangent to the parabola,$x^2 = 8y,$ which makes an angle $\theta$ with the positive direction of the $x-$axis,is

The distance of the point $(6, -2 \sqrt{2})$ from the common tangent $y = mx + c$ $(m > 0)$ of the curves $x = 2y^2$ and $x = 1 + y^2$ is

From the point $(-1, -6)$, two tangents are drawn to the parabola $y^2 = 4x$. The angle between the two tangents is:

If the vertex of the parabola $y = x^2 - 8x + c$ lies on the $x$-axis,then the value of $c$ is:

If $\overrightarrow{AB}$ is the focal chord of the parabola $y^2=16x$ and $A=(1,-4)$,then the equation of the normal to the parabola at the point $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module