If for a variable line frac{x}{a} + frac{y}{b | vedclass

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

Question

(C) The equation of the given line is $\frac{x}{a} + \frac{y}{b} = 1$ .....$(i)$The slope of this line is $m_1 = -\frac{b}{a}$.The slope of the perpen

Vedclass · Accepted Answer

$x^2 + y^2 = c^2$

Vedclass · Answer

(C) The equation of the given line is $\frac{x}{a} + \frac{y}{b} = 1$ .....$(i)$The slope of this line is $m_1 = -\frac{b}{a}$.The slope of the perpendicular line passing through the origin $(0, 0)$ is $m_2 = \frac{a}{b}$.The equation of the perpendicular line is $y - 0 = \frac{a}{b}(x - 0)$,which simplifies to $\frac{x}{b} - \frac{y}{a} = 0$ .....$(ii)$Let $(x, y)$ be the foot of the perpendicular. Squaring and adding equations $(i)$ and $(ii)$:$\left(\frac{x}{a} + \frac{y}{b}\right)^2 + \left(\frac{x}{b} - \frac{y}{a}\right)^2 = 1^2 + 0^2$$x^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) + y^2\left(\frac{1}{b^2} + \frac{1}{a^2}\right) = 1$Since $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,we substitute this into the equation:$(x^2 + y^2)\left(\frac{1}{c^2}\right) = 1$Therefore,the locus is $x^2 + y^2 = c^2$.

If for a variable line $\frac{x}{a} + \frac{y}{b} = 1$,the condition $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$ ($c$ is a constant) is satisfied,then the locus of the foot of the perpendicular drawn from the origin to the line is:

Similar Questions

$A$ point $P$ moves inside a square of area $4$ square units such that it is nearer to the point of intersection of its diagonals than any vertex. The area of the region traced by $P$ is

If $M$ is the foot of the perpendicular drawn from the origin $O$ to a variable line $L$ passing through a fixed point $Q(a, b)$,then the locus of the mid-point of $OM$ is

The Cartesian form of the polar equation $\theta = \tan^{-1} 2$ is

$A$ straight line passing through a fixed point $(2,3)$ intersects the coordinate axes at points $P$ and $Q$. If $O$ is the origin and $R$ is a variable point such that $OPRQ$ is a rectangle,then the locus of $R$ is

The perpendicular bisector of the line segment joining the points $P(1, 4)$ and $Q(k, 3)$ has a $y$-intercept of $-4$. Then a possible value of $k$ among the following is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module