The length of the chord of contact of the tan | vedclass

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

Question

(C) The equation of the chord of contact of tangents drawn from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.Here

Vedclass · Accepted Answer

$\frac{3}{2}\sqrt{41}$

Vedclass · Answer

(C) The equation of the chord of contact of tangents drawn from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.Here,the parabola is $y^2 = 8x$,so $4a = 8$,which means $a = 2$. The point is $(2, 5)$.Substituting these values,the equation of the chord of contact is $5y = 2 	imes 2(x + 2)$,which simplifies to $5y = 4x + 8$,or $x = \frac{5y - 8}{4}$.Substituting $x$ into the parabola equation $y^2 = 8x$:$y^2 = 8 \left( \frac{5y - 8}{4} ight) = 2(5y - 8) = 10y - 16$.$y^2 - 10y + 16 = 0$,which factors as $(y - 2)(y - 8) = 0$.Thus,$y = 2$ or $y = 8$.If $y = 2$,$x = \frac{5(2) - 8}{4} = \frac{2}{4} = 0.5$. If $y = 8$,$x = \frac{5(8) - 8}{4} = \frac{32}{4} = 8$.The points of intersection are $(0.5, 2)$ and $(8, 8)$.The length of the chord is $\sqrt{(8 - 0.5)^2 + (8 - 2)^2} = \sqrt{(7.5)^2 + 6^2} = \sqrt{56.25 + 36} = \sqrt{92.25} = \sqrt{\frac{369}{4}} = \frac{3}{2}\sqrt{41}$.

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

The length of the chord of contact of the tangents drawn from the point $(2, 5)$ to the parabola $y^2 = 8x$ is

Similar Questions

$P$ is a point on the parabola $y^2 = 4ax$ $(a > 0)$ whose vertex is $A$. $PA$ is produced to meet the directrix in $D$ and $M$ is the foot of the perpendicular from $P$ on the directrix. If a circle is described on $MD$ as a diameter,then it intersects the $x$-axis at a point whose coordinates are:

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

The equation of the tangent to the parabola $y^2 = 4x + 5$ parallel to the line $y = 2x + 7$ is

The points of intersection of the parabolas $y^2=5x$ and $x^2=5y$ lie on the line

Vedclass Test Series

Exam Paper Generator

Online Exam Module