If a variable line drawn through the point of | vedclass

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

Question

(B) The equation of a line passing through the intersection of the lines $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ and $\frac{x}{\beta} + \frac{y}{\alp

Vedclass · Accepted Answer

$\alpha \beta (x + y) = 2xy(\alpha + \beta)$

Vedclass · Answer

(B) The equation of a line passing through the intersection of the lines $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ and $\frac{x}{\beta} + \frac{y}{\alpha} = 1$ is given by the family of lines equation: $\left( \frac{x}{\alpha} + \frac{y}{\beta} - 1 \right) + \lambda \left( \frac{x}{\beta} + \frac{y}{\alpha} - 1 \right) = 0$. Rearranging this,we get $x \left( \frac{1}{\alpha} + \frac{\lambda}{\beta} \right) + y \left( \frac{1}{\beta} + \frac{\lambda}{\alpha} \right) = 1 + \lambda$. This line meets the axes at $A \left( \frac{1 + \lambda}{\frac{1}{\alpha} + \frac{\lambda}{\beta}}, 0 \right)$ and $B \left( 0, \frac{1 + \lambda}{\frac{1}{\beta} + \frac{\lambda}{\alpha}} \right)$. Let $(h, k)$ be the midpoint of $AB$. Then $h = \frac{1 + \lambda}{2(\frac{1}{\alpha} + \frac{\lambda}{\beta})}$ and $k = \frac{1 + \lambda}{2(\frac{1}{\beta} + \frac{\lambda}{\alpha})}$. From these,$\frac{1}{2h} = \frac{\frac{1}{\alpha} + \frac{\lambda}{\beta}}{1 + \lambda}$ and $\frac{1}{2k} = \frac{\frac{1}{\beta} + \frac{\lambda}{\alpha}}{1 + \lambda}$. Adding these,$\frac{1}{2h} + \frac{1}{2k} = \frac{\frac{1}{\alpha} + \frac{1}{\beta} + \lambda(\frac{1}{\alpha} + \frac{1}{\beta})}{1 + \lambda} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$. Thus,$\frac{h + k}{2hk} = \frac{\alpha + \beta}{\alpha \beta}$,which simplifies to $\alpha \beta (h + k) = 2hk(\alpha + \beta)$. The locus of $(h, k)$ is $\alpha \beta (x + y) = 2xy(\alpha + \beta)$.

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ and $\frac{x}{\beta} + \frac{y}{\alpha} = 1$ meets the coordinate axes in $A$ and $B$,then the locus of the midpoint of $AB$ is

Similar Questions

$A(1, 0)$,$B(0, 2)$,and $C(1, 2)$ are three points on the $XY$-plane. If a point $P(x, y)$ moves such that the area of $\triangle PAB$ is twice the area of $\triangle ABC$,then the locus of the point $P$ is:

$B(2,3)$,$C(5,-2)$,and $D(1,-1)$ are three points. If $A(x, y)$ is a variable point such that the area of the quadrilateral $ABCD$ is $10 \text{ sq. units}$,then the locus of $A$ is

$A$ variable line passes through a fixed point $(a, b)$ and meets the coordinate axes at $A$ and $B$. The locus of the point of intersection of lines drawn through $A$ and $B$ parallel to the coordinate axes is:

$A$ straight line cuts off intercepts on the coordinate axes,the reciprocal of whose sum is $1/p$. Through which fixed point does this line always pass?

$A$ variable line passing through $(l, m)$ intersects the coordinate axes at the points $A$ and $B$. If the line drawn parallel to $Y$-axis through $A$ and parallel to $X$-axis through $B$ meet at $P$,then the locus of $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module