Question

A chord of a parabola cuts the axis of the parabola at O. The feet of the perpendiculars from P and P’ on the axis are M and M’ respectively. If V is the vertex then VM, VO, VM’ are
(a) A.P
(b) G.P
(c) H.P
(d) AP, GP

Answer 1

Hint: To solve this question we will first take an equation of a parabola and then try to draw its figure using the given points in the question. Any point P on the parabola is of the form $$P(a{t}^2,2at)$$ where t varies. We will use that the slope of the line with endpoints is given by $${\displaystyle \frac{y_1-y_2}{x_1-x_2}}$$ where $$(x_1,y_1)$$ and $$(x_2,y_2)$$ are the endpoints.

Complete step-by-step answer:
To solve this question, we will first consider some parabola. To do that let us define a parabola and some examples. The parabola is the locus of points in that plane that are equidistant from both the directrix and the focus. Another description of a parabola is a conic section, created from the intersection of a right circular conical surface and a plane parallel to another plane that is tangential to the conical surface. The examples of the standard parabola are:
$$\left(I\right)y^2=x$$
It is drawn as

$$\left(II\right)y=x^2$$
It is drawn as

  From these, let the parabola be $$y^2=4ax$$

Let the vertex of parabola be V = (0, 0). The chord PP’ cuts x-axis at 0 and let it be (R, 0).
$$\Rightarrow VO=R$$
Let $$P=(a{t_1}^2,2a{t}_1)$$ be the coordinates of the point P. This is so as any point on the parabola is of the form $$P=(a{t}^2,2at)$$ where t varies. Then the foot of the perpendicular on the axis is $$M=(a{t_1}^2,0)$$ (as visible by the diagram) and similarly for $$P^{{}^{\prime }}=(a{t_2}^2,2a{t}_2)$$ the foot of the perpendicular is $$M^{{}^{\prime }}(a{t_2}^2,0)$$ the coordinate of the point (0, 0). The slope of the line having endpoints as $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is given by $${\displaystyle \frac{y_1-y_2}{x_1-x_2}}.$$ Then for PO, the slope is given by the slope of $$PO={\displaystyle \frac{2a{t}_1}{a{t_1}^2-R}}.$$
Similarly, the slope of P’O is given by (using the above formula) the slope of $$P^{{}^{\prime }}O={\displaystyle \frac{-2a{t}_2}{R-a{t_2}^2}}.$$
Now because PO and P’O are forming the same line, so then the slopes are equal.
The slope of PO = Slope of P’O
$$\Rightarrow {\displaystyle \frac{2a{t}_1}{a{t_1}^2-R}}={\displaystyle \frac{-2a{t}_2}{R-a{t_2}^2}}$$
On cross multiplying, we get,
$$\Rightarrow 2a{t}_1(R-a{t_2}^2)=-2a{t}_2(a{t_1}^2-R)$$
$$\Rightarrow t_{1}R-a{t_2}^2{t}_1=-a{t_1}^2{t}_2+t_{2}R$$
$$\Rightarrow R(t_1-t_2)=a{t_2}^2{t}_1-a{t_1}^2{t}_2$$
$$\Rightarrow R={\displaystyle \frac{a{t}_1{t}_2(t_2-t_1)}{(t_1-t_2)}}$$
$$\Rightarrow R=-a{t}_1{t}_2$$
Hence, the value of R is $$-a{t}_1{t}_2.$$
So, we have,
$$VO=R$$
$$VM=a{t_1}^2$$
$$V{M}^{{}^{\prime }}=a{t_2}^2$$
$$\Rightarrow V{O}^2=R^2$$
Substituting $$R=-a{t}_1{t}_2$$
$$\Rightarrow V{O}^2=R^2$$
$$\Rightarrow V{O}^2={(-a{t}_1{t}_2)}^2$$
$$\Rightarrow V{O}^2=a^2{t_1}^2{t_2}^2$$
$$\Rightarrow V{O}^2=a{t_1}^{2}a{t_2}^2$$
$$\Rightarrow V{O}^2=VM.V{M}^{{}^{\prime }}$$
$$\Rightarrow {\left(VO\right)}^2=VM.V{M}^{{}^{\prime }}$$
Hence, VM, VO, VM’ are in GP.

So, the correct answer is “Option (c)”.

Note: When three numbers a, b and c are in GP, then they can be written as $$ac=b^2.$$ Here, we have obtained the answer as $${\left(VO\right)}^2=VM.V{M}^{{}^{\prime }}$$ the number using the above stated theory are in GP.