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  • Stuck here, help me understand: - Co-ordinate geometry - JEE Main-5

The normal to a curve at P(x,y) meets the x-axis at G . If the distance of  G from the origin is twice the abscissa of P , then the curve is a

  • Option 1)

    circle

  • Option 2)

    hyperbola

  • Option 3)

    ellipse

  • Option 4)

    parabola

 

Answers (1)

best_answer

As we learnt in

Slope of a line -

m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

- wherein

Slope of line joining A(x1,y1) and  B(x2,y2) .

 

 If slope of tangent = m

We get  \frac{y-y_{1}}{x-x_{1}}=\frac{-1}{m}    as equation of normal.

If normal meets x-axis again, we get:

x=my_{1}+x_{1}

Distance of QG = 2 (abscissa of P)

i.e.,    my_{1}+x_{1}=2x_{1}

\Rightarrow m=\frac{x_{1}}{y_{1}}\:\:\:\:at\:\:\:\:(x_{1}y_{1})

Generalising  \frac{dy}{dx}=\frac{x}{y}

\frac{y^{2}}{2}=\frac{x^{2}}{2}+c

\frac{x^{2}}{2}-\frac{y^{2}}{2}=c    which is a Hyperbola.


Option 1)

circle

This option is incorrect.

Option 2)

hyperbola

This option is correct.

Option 3)

ellipse

This option is incorrect.

Option 4)

parabola

This option is incorrect.

Posted by

divya.saini

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