Question

Tangent drew on ellipse 2x^2+y^2 =1 at any point (not in vertex) cut axis of coordinate at point A & B. then locus of midpoint of AB is?
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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

The given equation of the ellipse is$2 {x}^{2} + {y}^{2} = 1 \: \: ...(1)$

Let ( h, k) be an arbitrary on the ellipse other than vertex

Then

$2 {h}^{2} + {k}^{2} = 1 \: \: \: ...(2)$

Differentiating both sides of Equation (1) with respect to x

$\displaystyle \: 4x + 2y \frac{dy}{dx} = 0$

$\implies \: \displaystyle \: \frac{dy}{dx} = - \frac{2x}{y}$

$\: \: \displaystyle \: \frac{dy}{dx} \left \right|_{( h, k)} = - \frac{2h}{k}$

So the equation of the tangent at ( h , k) is$\: \: \displaystyle \: (y - k)= - \frac{2h}{k} (x - h)$

$\implies \: \: \: \displaystyle \: (ky - {k}^{2} )= -2xh + 2 {h}^{2}$

$\implies \: \: \: \displaystyle \: 2xh + yk = 2 {h}^{2} + {k}^{2}$

$\implies \: \: \: \displaystyle \: 2xh + yk = 1 \: \: \: \: (\: by \: equation\: 2)$

Which can be rewritten as

$\implies \: \: \: \displaystyle \: \frac{x}{ \frac{1}{2h} } + \frac{y}{ \frac{1}{k} } = 1$

So the straight line intersect coordinate axis at

$\displaystyle \: A \: ( \frac{1}{2h} ,0) \: \: and \: \: B \: ( 0 \: , \: \frac{1}{k} )$

Let (u, v) be the middle point of AB

SO$\displaystyle \: u = \frac{ \frac{1}{2h} + 0}{2} = \frac{1}{4h}$

$v = \displaystyle \: \frac{ 0 + \frac{1}{k} }{2} = \frac{1}{2k}$

So

$\displaystyle \: h = \frac{1}{4u}$

$\displaystyle \: k = \frac{1}{2v}$

From Equation (2)

$\displaystyle \: 2 \times { \bigg( \frac{1}{4u} \bigg) }^{2} + { \bigg( \frac{1}{2v} \bigg) }^{2} \ = 1$

$\implies \: \displaystyle \: \frac{1}{8 {u}^{2} } + \frac{1}{ 4{v}^{2} } = 1$

RESULT

So the required locus is

$\displaystyle \: \frac{1}{8 {x}^{2} } + \frac{1}{ 4{y}^{2} } = 1$