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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 (2)

best_answer

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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