Test Index

VITEEE 2013 Paper

Section: Mathematics

Question : 109 of 120

Marks: +1, -0

\frac{x^2}{a^2} + \frac{y^2}{b^2}

= 1 will be

Solution:

Let mid-point of part PQ which is in between the axis is R

(x_1,y_1)

, then coordinates of P and Q will be

(2x_1,0)

and

(0,2y_1)

respectively.

∴ Equation oflinePQis

\frac{x}{2x_1} + \frac{y}{2y_1}

= 1

⇒ y = -

\left(\frac{y_1}{x_1}\right)

x + 2

y_1

If this line touches the ellipse

\frac{x^2}{a^2} + \frac{y^2}{b^2}

= 1

then it will satisfy the condition,

x^2

a^2 m^2 + b^2

So,

(2y_1)^2

a^2 \left(\frac{-y_1}{x_1}\right)^2 + b^2

⇒

4y_1^2

\left\{\frac{a^2 y_1^2}{x_1^2}\right\} + b^2

⇒ 4 =

\left(\frac{a^2}{x_1^2}\right) + \left(\frac{b^2}{y_1^2}\right)

⇒

\left(\frac{a^2}{x_1^2}\right)+\left(\frac{b^2}{y_1^2}\right)

= 4

∴ Required locus of

(x_1 , y_1)

\frac{a^2}{x^2} + \frac{b^2}{y^2}

= 4

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