If two distinct chords drawn from the point A | vedclass

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

Question

(A) Let the point of intersection of the chord with the line $y=ax$ be $M(\alpha, a\alpha)$. Since $M$ is the midpoint of the chord with one endpoint

Vedclass · Accepted Answer

$\left(\frac{1}{2}-\frac{1}{\sqrt{2}}, \frac{1}{2}+\frac{1}{\sqrt{2}}\right)$

Vedclass · Answer

(A) Let the point of intersection of the chord with the line $y=ax$ be $M(\alpha, a\alpha)$. Since $M$ is the midpoint of the chord with one endpoint $A(4,4)$ and the other endpoint $Q(x_1, y_1)$,we have:$\alpha = \frac{4+x_1}{2} \Rightarrow x_1 = 2\alpha - 4$$a\alpha = \frac{4+y_1}{2} \Rightarrow y_1 = 2a\alpha - 4$Since $Q(x_1, y_1)$ lies on the parabola $y^2=4x$,we substitute the coordinates:$(2a\alpha - 4)^2 = 4(2\alpha - 4)$$4a^2\alpha^2 - 16a\alpha + 16 = 8\alpha - 16$$4a^2\alpha^2 - (16a+8)\alpha + 32 = 0$For two distinct chords,the quadratic equation in $\alpha$ must have two distinct real roots,so the discriminant $D > 0$:$D = (16a+8)^2 - 4(4a^2)(32) > 0$$64(2a+1)^2 - 512a^2 > 0$$64(4a^2 + 4a + 1) - 512a^2 > 0$$256a^2 + 256a + 64 - 512a^2 > 0$$-256a^2 + 256a + 64 > 0$$256a^2 - 256a - 64 $4a^2 - 4a - 1 Solving $4a^2 - 4a - 1 = 0$ gives $a = \frac{4 \pm \sqrt{16 - 4(4)(-1)}}{8} = \frac{4 \pm \sqrt{32}}{8} = \frac{1 \pm \sqrt{2}}{2}$.Thus,the interval is $\left(\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right)$,which is $\left(\frac{1}{2}-\frac{1}{\sqrt{2}}, \frac{1}{2}+\frac{1}{\sqrt{2}}\right)$.

If two distinct chords drawn from the point $A(4,4)$ on the parabola $y^2=4x$ are bisected by the line $y=ax$,then the interval in which $a$ lies is

Similar Questions

If all the vertices of an equilateral triangle lie on the parabola $y^2=16x$ and one of them coincides with the vertex of that parabola,then the length of the side of that triangle is

$A$ chord is drawn through the focus of the parabola $y^2 = 6x$ such that its distance from the vertex of this parabola is $\frac{\sqrt{5}}{2}$. Then,its slope can be:

$A$ point $P$ moves such that the sum of the angles which the three normals drawn from $P$ to the standard parabola $y^2 = 4ax$ make with the axis of the parabola is constant. Then the locus of $P$ is:

$x - 2 = t^2$ and $y = 2t$ are the parametric equations of which parabola?

Tangents drawn from the point $(-8, 0)$ to the parabola $y^2 = 8x$ touch the parabola at $P$ and $Q$. If $F$ is the focus of the parabola,then the area of the triangle $PFQ$ (in sq. units) is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module