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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

 
Answers (1)
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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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