Q

# Can someone explain - Co-ordinate geometry - JEE Main-5

A normal to the hyperbola, $4x^{2}-9y^{2}=36$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the locus of P is :

• Option 1)

$4x^{2}+9y^{2}=121$

• Option 2)

$9x^{2}+4y^{2}=169$

• Option 3)

$4x^{2}-9y^{2}=121$

• Option 4)

$9x^{2}-4y^{2}=169$

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As we learned,

Locus -

Path followed by a point p(x,y) under given condition (s).

- wherein

It is satisfied by all the points (x,y) on the locus.

and

Mid-point formula -

$x= \frac{x_{1}+x_{2}}{2}$

$y= \frac{y_{1}+y_{2}}{2}$

- wherein

If the point P(x,y) is the mid point of line joining A(x1,y1) and B(x2,y2) .

Given Hyperbola $\Rightarrow \: 4x^{2}-9y^{2}=36$

$\left ( x_{o},y_{o} \right )$ be point of contact.

Slope of normal = $-\frac{9y}{4x}$

equation $\Rightarrow \: y-y_{o}=-\frac{9y_{o}}{4x_{o}}\left ( x-x_{o} \right )$

line intersects X axis at A when y = 0

$A=\left ( \frac{13x_{o}}{9},0 \right )\! B\left ( 0,\frac{13y_{o}}{4} \right )$

midpoint of OB = $B\left ( 0,\frac{13y_{o}}{8} \right )$

midpoint of AP = $\left ( \frac{\frac{13x_{x}}{9}+x}{9} ,\frac{y}{2}\right )$

Thus $x_{o}=-\frac{9x}{13},\: y=\frac{4y}{13}$

Put in Hyperbola , we get

$9x^{2}-4y^{2}=169$

Option 1)

$4x^{2}+9y^{2}=121$

Option 2)

$9x^{2}+4y^{2}=169$

Option 3)

$4x^{2}-9y^{2}=121$

Option 4)

$9x^{2}-4y^{2}=169$

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