Name: Pair of Tangent, Chord of Contact
Uploaded: 2019-07-19
Duration: 19 min 37 s
Description: This video contains all the details of the chord of the contact pair of tangent and cord bisected at a point. This concept is very useful for the students.

Question

Parabolas are drawn to touch two given rectangular axes and their foci are all at a constant distance

c

from the origin. Prove that the locus of the vertices of these parabolas is the curve

x^{\frac{2}{3}} + y^{\frac{2}{3}} = c^{\frac{2}{3}}

.

Accepted Answer

Let the vertex $V$ be $(h, k)$ .

Since focus $S$ is at constant distance $c$ from the origin $O$ , it lies on a circle with center at origin and radius $c$ with equation $x^{2} + y^{2} = c^{2}$ .

Let its coordinates be $(c \cos θ, c \sin θ)$ .

Since $V$ is the mid-point of $M S$ , where $M$ is the point of intersection of axis and directrix of the parabola, we get coordinates of $M (2 h - c \cos θ, 2 k - c \sin θ) . . . (i)$ (Using mid-point formula).

Now, recall that from any point on the directrix of the parabola, perpendicular tangents can be drawn to it (as directrix is the director circle of the parabola).

Hence, directrix passes through the origin (since from origin, two perpendicular tangents, and the coordinate axes are drawn).

So, slope of directrix $=$ slope of $O M = \frac{2 k - c \sin θ}{2 h - c \cos θ} . . . (ii)$

Slope of axis $=$ slope of $S V = \frac{c \sin θ - k}{c \cos θ - h} . . . (iii)$

Since axis and directrix are always perpendicular, so

$\frac{2 k - c \sin θ}{2 h - c \cos θ} \times \frac{c \sin θ - k}{c \cos θ - h} = - 1$

$\Rightarrow 3 c (k \sin θ + h \cos θ) = 2 h^{2} + 2 k^{2} + c^{2}$

Dividing by $c^{2}$ , we get

$3 [(\frac{k}{c}) \sin θ + (\frac{h}{c}) \cos θ] = 2 [{(\frac{h}{c})}^{2} + {(\frac{k}{c})}^{2}] + 1 . . . (iv)$

Now consider,

$2 (\cos^{6} θ + \sin^{6} θ) + 1 = 2 ({(\cos^{2} θ + \sin^{2} θ)}^{3} - 3 \cos^{2} θ \sin^{2} θ (\cos^{2} θ + \sin^{2} θ)) + 1$

$= 2 (1 - 3 \cos^{2} θ \sin^{2} θ) + 1 = 3 (1 - 2 \sin^{2} θ \cos^{2} θ)$

$= 3 ({(\sin^{2} θ + \cos^{2} θ)}^{2} - 2 \sin^{2} θ \cos^{2} θ) = 3 (\sin^{4} θ + \cos^{4} θ)$

i.e., $3 (\sin^{4} θ + \cos^{4} θ) = 2 (\cos^{6} θ + \sin^{6} θ) + 1 . . . (v)$

Now comparing with $(iv)$ , we get $\frac{k}{c} = \sin^{3} θ, \frac{h}{c} = \cos^{3} θ$ .

$\Rightarrow \sin θ = {(\frac{k}{c})}^{\frac{1}{3}}, \cos θ = {(\frac{h}{c})}^{\frac{1}{3}}$

To find the required locus, we need to eliminate $θ$ .

So using $\sin^{2} θ + \cos^{2} θ = 1$ ,

$\Rightarrow {(\frac{k}{c})}^{\frac{2}{3}} + {(\frac{h}{c})}^{\frac{2}{3}} = 1$

$\Rightarrow {(k)}^{\frac{2}{3}} + {(h)}^{\frac{2}{3}} = {(c)}^{\frac{2}{3}}$

Generalizing the locus, we get $x^{\frac{2}{3}} + y^{\frac{2}{3}} = c^{\frac{2}{3}}$ .

Hence proved.

Parabolas are drawn to touch two given rectangular axes and their foci are all at a constant distance $c$ from the origin. Prove that the locus of the vertices of these parabolas is the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = c^{\frac{2}{3}}$ .

Important Questions on The Parabola (Continued)

Learn and Prepare for any exam you want !

Parabolas are drawn to touch two given rectangular axes and their foci are all at a constant distance c from the origin. Prove that the locus of the vertices of these parabolas is the curve x23+y23=c23.

Important Questions on The Parabola (Continued)

Learn and Prepare for any exam you want !

Parabolas are drawn to touch two given rectangular axes and their foci are all at a constant distance $c$ from the origin. Prove that the locus of the vertices of these parabolas is the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = c^{\frac{2}{3}}$ .