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1 week, 5 days ago

Can someone explain - Co-ordinate geometry - JEE Main-6

 The eccentricity of an ellipse whose centre is at the origin is \frac{1}{2}

 If one of its directrices  is x=−4, then the equation of the normal to it at

\left ( 1,\frac{3}{2} \right ) is:

 

  • Option 1)

    4x-2y=1

  • Option 2)

    4x+2y=7

  • Option 3)

    x+2y=4

  • Option 4)

    2y-x=2

 
Answers (1)
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S solutionqc
Answered 1 week, 5 days ago

As learned in concept

Eccentricity -

e= \sqrt{1-\frac{b^{2}}{a^{2}}}

- wherein

For the ellipse  

\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1

 

 and

Equation of directrices -

x= \pm \frac{a}{e}

- wherein

For the ellipse  

\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1

 

 

we have e= \frac{1}{2} &  x=-4

 

From this we can calculate the value of a

-\frac{a}{e}=-4 \Rightarrow a=4e

 so a=4

Now,  b^{2}=a^{2}(1-e^{2})=3

Hence, Equation of ellipse is-

\frac{x^{2}}{4}+\frac{y^{2}}{3}=1

Differentiate to get the value of slope- 

\frac{x}{2}+\frac{2y}{3}*y^{'}=0 \Rightarrow y^{'}=\frac{3x}{4y}

So slope at given point     (1'\frac{3}{2}) is -\frac{1}{2}

Now by using the formula of normal of a equation-

y-y_{1}=-\frac{1}{m}(x-x_{1})   normal at point   (1'\frac{3}{2}) is   y-\frac{3}{2}=2*(x-1)

we get the equation of the normal to be    4x-2y=1

 


Option 1)

4x-2y=1

Correct

Option 2)

4x+2y=7

Incorrect

Option 3)

x+2y=4

Incorrect

 

 

Option 4)

2y-x=2

Incorrect

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