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2 months, 3 weeks ago

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

The eccentricity of an ellipse whose centre is at the origin is

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

is:

• Option 1)

• Option 2)

• Option 3)

• Option 4)

73 Views
N Neha Gupta
S solutionqc
Answered 2 months, 3 weeks 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)

Correct

Option 2)

Incorrect

Option 3)

Incorrect

Option 4)

Incorrect

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