Let x=4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 12. If P (1 , ) , 0 is a point on this ellipse, then the equation of the normal to it at P is

Question

Let x=4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 12. If P (1 , ) , >0 is a point on this ellipse, then the equation of the normal to it at P is

Accepted Answer

Correct answer: B. (b) JEE Main 2020 Given x=4 ⇒ ae=4 ⇒a=4e ⇒a=4 12=2 ⇒a=2 Now, b^2=a^2(1-e^2) ⇒b^2=4(1- 14)=4 34=3 So, equation x^24+ y^23=1 ⇒3x^2+4y^2=12 Now, P (1 , ) lies on it ⇒3+4 ^2=12 ⇒4 ^2=9 ⇒ = 32 So, P (1 , 32 ) Now differentiating (i) 6x+8y dy dx=0 ⇒ dy dx= -3x4y ⇒ ( dy dx ) (1 , 32 )=- 12 So, slope of normal =2 So, equation of normal : y- 32=2(x-1) ⇒2y-3=4x-4 ⇒4x-2y=1

$Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is \frac{1}{2} . If P (1, β), β > 0 is a point on this ellipse, then the equation of the normal to it at P is$

Solution

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