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

A parabola is drawn touching the

x

-axis at the origin and having its vertex at a given distance

k

from this axis. Prove that the axis of the parabola is a tangent to the parabola

x^{2} = - 8 k (y - 2 k)

.

Accepted Answer

Let the equation to any parabola in its most general form be given as,

${(a x + b y)}^{2} + 2 g x + 2 f y + c = 0 . . . (i)$

As the parabola passes through the origin (which means $c$ is $0$ ) and touches the $x$ -axis, (which means $y$ is $0$ ), we get $y = 0$ is a tangent.

So, substituting the value of $y$ as $0$ in equation $(i)$ , we get,

$a^{2} x^{2} + 2 g x = 0$

If $y = 0$ is a tangent, this equation must be a perfect square, as it just touches at one point (which means repeated roots of the equation in $x$ ). So, we get,

$g = 0$

Substituting the values of $c$ and $g$ in equation $(i)$ , we get,

${(a x + b y)}^{2} + 2 f y = 0 . . . (i i)$

The axis of $(i i)$ can be given as,

$a x + b y = - \frac{b f}{(a^{2} + b^{2})} . . . (i i i)$

To find out the vertex, we need to solve equations $(i i)$ and $(i i i)$ , simultaneously.

Substituting the value of $a x + b y$ from equation $(i i i)$ in equation $(i i)$ , we get,

$\frac{b^{2} f^{2}}{{(a^{2} + b^{2})}^{2}} + 2 f y = 0$

$\Rightarrow y = - \frac{b^{2} f}{2 {(a^{2} + b^{2})}^{2}}$

So, the ordinate of the vertex is given as,

$- \frac{b^{2} f}{2 {(a^{2} + b^{2})}^{2}} = k$ (by hypothesis)

$\Rightarrow f = - \frac{2 k {(a^{2} + b^{2})}^{2}}{b^{2}}$

Substituting the value of $f$ in equation $(i i i)$ , we get,

$a x + b y = b \times \frac{2 k {(a^{2} + b^{2})}^{2}}{b^{2} (a^{2} + b^{2})}$

$a x + b y = \frac{2 k (a^{2} + b^{2})}{b}$

$x = - \frac{b}{a} y + \frac{2 k}{a b} (a^{2} + b^{2})$

$x = - \frac{b}{a} + 2 k (\frac{a}{b} + \frac{b}{a}) . . . (i v)$

The equation of any tangent to the given parabola can be given as,

$x^{2} = - 8 k (y - 2 k)$

$x^{2} = 4 . (- 2 k) (y - 2 k)$

We can write the above equation of the tangent as,

$x = m (y - 2 k) - \frac{2 k}{m}$ $(∴ a = - 2 k)$

$x = m y - 2 k m - \frac{2 k}{m}$

$x = m y - 2 k (m + \frac{1}{m}) . . . (v)$

As it is true to any value of $m$ , substituting $m$ as $- \frac{b}{a}$ in equation $(v)$ , we get,

$x = \frac{b}{a} y - 2 k (- \frac{b}{a} - \frac{a}{b})$

$x = - \frac{b}{a} y + 2 k (\frac{b}{a} + \frac{a}{b})$ (which is same as equation $(i v)$ , that is, the axis of the parabola)

Hence proved.

A parabola is drawn touching the $x$ -axis at the origin and having its vertex at a given distance $k$ from this axis. Prove that the axis of the parabola is a tangent to the parabola $x^{2} = - 8 k (y - 2 k)$ .

Important Questions on Conic Sections. The Parabola

Learn and Prepare for any exam you want !

A parabola is drawn touching the x-axis at the origin and having its vertex at a given distance k from this axis. Prove that the axis of the parabola is a tangent to the parabola x2=-8ky-2k.

Important Questions on Conic Sections. The Parabola

Learn and Prepare for any exam you want !

A parabola is drawn touching the $x$ -axis at the origin and having its vertex at a given distance $k$ from this axis. Prove that the axis of the parabola is a tangent to the parabola $x^{2} = - 8 k (y - 2 k)$ .