The ends $$A, B$$ of a straight line segment of constant length $$c$$ slide upon the fixed rectangular axes $$OX$$ & $$OY$$ respectively. If the rectangle $$OAPB$$ be completed then, the locus of the foot of the perpendicular drawn from $$P$$ to $$AB$$ is ?
Let $$OA = a$$ and $$OB = b$$. Then, the co-ordinates of $$A$$ and $$B$$ are $$(a, 0)$$ and $$(0, b)$$ respectively.
The co-ordinates of $$P$$ are $$(a, b).$$ Let $$\theta$$ be the foot of perpendicular from $$P$$ on $$AB$$ and let the coordinates of $$Q(h, k).$$
Here $$a$$ and $$b$$ are the variable and we have to find locus of $$Q.$$
Now, $$AB = c$$
$$\Rightarrow AB^2=c^2\Rightarrow OA^2+OB^2=c^2$$
$$\Rightarrow a^2+b^2+c^2$$ .......(1)
Now $$PQ\perp AB$$
$$\Rightarrow$$ Slope of $$AB.$$ Slope of $$PQ=-1$$
$$\Rightarrow \dfrac {k-b}{h-a}\cdot \dfrac {0-b}{a-0}=-1$$
$$\Rightarrow bk-b^2=ah-a^2$$
$$\Rightarrow ah-bk=a^2-b^2$$ .......(2)
Equation of line $$AB$$ is $$\dfrac {x}{a}+\dfrac {y}{b}=1$$
Since, $$Q$$ lies on $$AB,$$
$$\dfrac {h}{a}+\dfrac {k}{b}=1\Rightarrow bh+ak=ab$$ .......(3)
Solving (2) and (3), we get
$$\dfrac {h}{ab^2+a(a^2-b^2)}=\dfrac {k}{-b(a^2-b^2)+a^2b}=\dfrac {1}{a^2+bb^2}$$
$$\Rightarrow \dfrac {h}{a^3}=\dfrac {k}{b^2}=\dfrac {1}{c^2}$$ {using (1)}
$$\Rightarrow a=(hc^2)^{1/3}$$ and $$b=(kc^2)^{1/3}$$
Substituting the values of $$a$$ and $$b$$ in $$a^2+b^2+c^2$$, we get
$$h^{2/3}+k^{2/3}=c^{2/3}$$
$$x^{2/3}+y^{2/3}=c^{2/3}$$ required locus.
Let $$a$$ and $$b$$ be any two numbers satisfying $$\dfrac {1}{a^2}+\dfrac {1}{b^2}=\dfrac {1}{4}$$. Then, the foot of perpendicular from the origin on the variable line, $$\dfrac {x}{a}+\dfrac {y}{b}=1$$, lies on :
The foot of the perpendicular drawn from the origin, on the line, $$3x + y = \lambda (\lambda \neq 0)$$ is $$P$$. If the line meets x-axis at $$A$$ and y-axis at $$B$$, then the ratio $$BP : PA$$ is
The foot of the perpendicular from the point $$(3, 4)$$ on the line $$3x-4y+5=0$$ is:
A straight line passes through a fixed point $$(h,k)$$. Find the locus of feet of the perpendiculars, drawn from the origin to it.
If $$p$$ and $$p'$$ be the lengths of perpendiculars from origin to the lines $$x \sec \theta-y \cos\theta=a$$ and $$x \cos \theta-y \sin \theta=a \cos 2\theta$$ respectively, then prove that $$4p^{ 2 }+p' ^{ 2 }=a^{ a }$$.
Prove that the product of the perpendiculars from the point $$ \left[ \pm \sqrt { \left( { a }^{ 2 }-{ b }^{ 2 } \right) } ,0 \right]$$ to the line $$\dfrac{ x }{ a }\cos \theta+\dfrac{ y }{ b }\sin \theta=1$$ is $$b^{ 2 }$$.
If $$x \cos \alpha + y \sin \alpha = p$$ where, $$p=\dfrac{ \sin^{ 2 }\alpha }{ \cos \alpha }$$ be a straight line, prove that perpendiculars $$p_{ 1 },p_{ 2 }$$ and $$ p_{ 3 }$$ on this line from the points $$(m^{ 2 }, 2m), (mm,m+m)$$ and $$(m^{ 2 },2m)$$ respectively are in geometrical progression.
Find the co-ordinates of the foot of the perpendicular drawn from the point$$(2,3)$$to the line $$y=3x+4$$
Let $$\triangle ABC$$ be a triangle with $$AB = AC$$. If $$D$$ is the midpoint of $$BC$$, $$E$$ is the foot of the perpendicular drawn from $$D$$ to $$AC$$ and $$F$$ the mid-point of $$DE$$. Then $$AF$$ is perpendicular to
The vertices of a triangle OBC are 0(0,0), B (-3, -1), C (- 1,- 3). Find the equation of the line parallel to BC and intersecting the sides OB and OC and whose perpendicular distance from the point (0 , 0) is 1/2.